JEE-Mathematics Paragraph for Question Nos. 10 to 12 Consider the functions defined implicitly by the equation y3 – 3y + x = 0 on various intervals in the real line. If x (–, –2)(2, ), the equation implicitly defines a unique real valued differentiable function y = f(x). If x (–2, 2), the equation implicitly defines a unique real valued differentiable function y = g(x) satisfying g(0) = 0. 1 0 . If f(–10 2 ) = 2 2 , then f''(–10 2 ) = [JEE 2008, (4M, –1M)] 42 42 42 42 (A) 7332 (B) 7332 (C) 733 (D) 733 1 1 . The area of the region bounded by the curve y = f(x), the x-axis, and the lines x = a and x = b, where – < a < b < –2, is - [JEE 2008, (4M, –1M)] bx bx dx bf(b) af(a) (A) a 3((f(x))2 1) dx bf(b) af(a) a3 (B) f x2 1 bx bx dx bf(b) af(a) (C) (D)– dx bf(b) af(a) a3 f x2 1 f x2 1 a3 1 [JEE 2008, (4M, –1M)] 1 2 . g '(x) dx (B) 0 (C) –2g(1) (D) 2g(1) 1 (A) 2g(–1) 13 . The area of the region between the curves y = 1 sin x and y = 1 sin x bounded by the lines cos x cos x [JEE 2008, (3M, –1M)] x = 0 and x = 4 is :- 2 1 t 2 1 4 t (A) dt (B) dt 0 (1 t2 ) 1 t2 0 (1 t2 ) 1 t2 2 1 4 t 2 1 t (C) dt (D) dt 0 (1 t2 ) 1 t2 0 (1 t2 ) 1 t2 NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#08\\Eng\\01. Area Under the curve.p65 1 4 . Area of the region bounded by the curve y = ex and lines x = 0 and y = e is - [JEE 2009, (4M, –1M)] (A) e – 1 e 1 e (B) ln(e 1 y)dy (C) e exdx (D) ln y dy 1 0 1 Paragraph for Question 15 to 17 [JEE 10, (3M each), –1M] Consider the polynomial f(x) = 1 + 2x + 3x2 + 4x3. Let s be the sum of all distinct real roots of f(x) and let t = |s|. 1 5 . The real number s lies in the interval (A) 1 , 0 (B) 11, 3 (C) 3 , 1 (D) 0, 1 4 4 4 2 4 1 6 . The area bounded by the curve y = f(x) and the lines x = 0, y = 0 and x = t, lies in the interval (A) 3 , 3 (B) 21 , 11 (C) (9, 10) (D) 0, 21 4 64 16 64 21 E
JEE-Mathematics 1 7 . The function f'(x) is (A) increasing in t, 1 and decreasing in 1 , t 4 4 (B) decreasing in t, 1 and increasing in 1 , t 4 4 (C) increasing in (–t, t) (D) decreasing in (–t, t) 1 8 . ( a ) Let the straight line x = b divide the area enclosed by y = (1 – x)2, y = 0 and x = 0 into two parts R1 (0 x b) d and R2 (b x 1) such that R1 R2 1 . Then b equals 4 3 1 1 1 (A) (B) (C) 3 (D) 4 2 4 (b) Let ƒ:[–1,2] [0,) be a continuous function such that ƒ(x) = ƒ(1–x) for all x [–1,2]. 2 Let R1 x ƒ(x)dx , and R2 be the area of the region bounded by y = ƒ (x), x=–1, x=2, and the 1 x-axis. Then - (A) R1 = 2R2 (B) R1 = 3R2 (C) 2R1 = R2 (D) 3R1 = R2 [JEE 2011, 3+3M] 19. The area enclosed by the curve y = sinx + cosx and y = |cosx – sinx| over the interval 0, is 2 [JEE(Advanced) 2013, 2M] (A) 4( 2 1) (B) 2 2 ( 2 1) (C) 2( 2 1) (D) 2 2 ( 2 1) PREVIOUS YEARS QUESTIONS ANSWER KEY EXERCISE-5 [B] NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#08\\Eng\\01. Area Under the curve.p65 1. C 2. B 3. 20 4 2 sq. units 4. (a) C; (b) D 5. (a) A; (b) A 6. D 3 1 8. 125 9 . (A) (p), (B) (s), (C) (p), (D) (r) 10. B 7. 3 sq. units sq. units B, C, D 1 5 . C 1 6 . A 1 7 . B 11. A 12. D 1 8 . (a) B; (b) C 3 13. B 14. 19. B 22 E
JEE-Mathematics BINOMIAL THEOREM 1. BINOMIAL EXPRESSION : Any algebraic expression which contains two dissimilar terms is called binomial expression. 11 1 For example : x – y, xy + , 1, 3 etc. x z (x y )1 / 3 2. BINOMIAL THEOREM : The formula by which any positive integral power of a binomial expression can be expanded in the form of a series is known as BINOMIAL THEOREM. n 0 1 nCr xnr yr If x, y R and n N, then : ( x + y)n = nC xn + nC xn-1 y + nC xn-2 y2 + ..... + nC xn-r yr + ..... + nC yn = 2 rn r 0 This theorem can be proved by induction. Observations : ( a ) The number of terms in the expansion is ( n+1) i.e. one more than the index. ( b ) The sum of the indices of x & y in each term is n. ( c ) The binomial coefficients of the terms (nC , nC .....) equidistant from the beginning and the end are equal. 01 i.e. nC = nC r r –1 (d) Symbol nC can also be denoted by n , C(n, r) or A n . r r r Some important expansions : (i) (1 + x)n = nC + nC x + nC x2 + ........ + nC xn. 01 2 n (ii) (1 – x)n = nC – nC x + nC x2 + ........ + (–1)n . nC xn. 01 2 n Note : The coefficient of xr in (1+x)n = nC & that in (1–x)n = (–1)r .nC rr Illustration 1 : Expand : (y + 2)6. Solution : 6C0y6 + 6C1y5.2 + 6C2y4.22 + 6C3y3.23 + 6C4y2. 24 + 6C5y1 . 25 + 6C6 . 26. = y6 + 12y5 + 60y4 + 160y3 + 240y2 + 192y + 64. Illustration 2 : Write first 4 terms of 2y2 7 1 5 NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#11\\Eng\\01.BINOMIAL.p65 Solution : 7C0, 7C1 2y2 , 7 C2 2y2 2 , 7C3 2y2 3 5 5 5 Illustration 3 : 183 73 3.18.7.25 The value of 36 6.243.2 15.81.4 20.27.8 15.9.16 6.3.32 64 is - (A) 1 (B) 2 (C) 3 (D) 4 Solution : The numerator is of the form a3 b3 3ab a b a b3 E Where, a = 18 and b = 7 Nr = (18 + 7)3 = (25)3 Denominator can be written as 36 6 C1 .35 .21 6 C 2 .34 .22 6 C 3 .33 .23 6 C 4 32 .24 6 C 5 3 .25 6 C 6 26 3 2 6 56 25 3 Nr (25)3 1 Ans. Dr (25)3 1
JEE-Mathematics Illustration 4 : If in the expansion of (1 + x)m (1 – x)n, the coefficients of x and x2 are 3 and – 6 respectively then Solution : m is - [JEE 99] (A) 6 (B) 9 (C) 12 (D) 24 (1 + x)m (1 – x)n = m x (m )(m 1).x2 1 nx n(n 1) x2 ...... 1 2 ...... 2 Coefficient of x = m – n = 3 ........(i) Coefficient of x2 = –mn + n(n 1) m (m 1) 6 ........(ii) 22 Solving (i) and (ii), we get m = 12 and n = 9. Do yourself - 1 : (i) Expand 3 x 2 x 5 (ii) Expand (y + x)n 2 Pascal's triangle : A triangular arrangement of numbers as shown. The 1 numbers give the coefficients for the expansion of (x + y)n. The first row 11 is for n = 0, the second for n = 1, etc. Each row has 1 as its first and 121 last number. Other numbers are generated by adding the two numbers 1331 immediately to the left and right in the row above. 14 6 41 1 5 10 10 5 1 1 6 15 20 15 6 1 etc. 3 . IMPORTANT TERMS IN THE BINOMIAL EXPANSION : ( a ) General term: The general term or the ( r +1)th term in the expansion of (x + y)n is given by T = nC x n–r yr r +1 r Illustration 5 : Find : (a) The coefficient of x7 in the expansion of a x 2 1 11 bx (b) The coefficient of x–7 in the expansion of ax 1 11 bx2 Also, find the relation between a and b, so that these coefficients are equal. NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#11\\Eng\\01.BINOMIAL.p65 Solution : (a) In the expansion of a x 2 1 11 , the general term is : bx 1 r a11r bx br Tr + 1 = 1 1 C r( a x 2) 11–r = 11 Cr . . x22 3r putting 22 – 3r = 7 3r = 15 r = 5 T6 = 11 C5 a6 .x7 b5 Hence the coefficient of x7 in a x 2 1 11 is 11C5a6b–5. Ans. bx E Note that binomial coefficient of sixth term is 11C5. 2
JEE-Mathematics (b) In the expansion of ax 1 11 , general term is : bx2 1 r a11r bx2 br Tr + 1 = 11Cr(ax)11–r = (–1)r 11Cr . x113r putting 11 – 3r = –7 3r = 18 r = 6 T7 = (–1)6. 11C6 a5 . x 7 b6 coefficient of x–7 in ax 1 11 11C6a5b–6. Hence the bx2 is Ans. Ans. Also given : Coefficient of x7 in a x 2 1 11 = coefficient of x–7 in ax 1 11 bx bx2 11C5a6b–5 =11C6a5b–6 ab = 1 ( 11C5 =11C6) which is the required relation between a and b. Illu str ation 6 : Find the number of rational terms in the expansion of (91/4 + 81/6) .1000 Solution : The general term in the expansion of (91/4 + 81/6)1000 is 1000r r Tr + 1 1 1 = C1 0 0 0 = C1 0 0 0 1000r r r 94 86 r 3 2 22 The above term will be rational if exponents of 3 and 2 are integers 1000 r r It means 2 and 2 must be integers The possible set of values of r is {0, 2, 4, ..........., 1000} Hence, number of rational terms is 501 Ans. NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#11\\Eng\\01.BINOMIAL.p65 (b) Middle term : The middle term(s) in the expansion of (x + y)n is (are) : ( i ) If n is even, there is only one middle term which is given by T = nC . x n/2. yn/2 (n +2)/2 n/2 ( i i ) If n is odd, there are two middle terms which are T (n +1)/2 &T [(n+1)/2]+1 Important Note : Middle term has greatest binomial coefficient and if there are 2 middle terms their coefficients will be equal. When r= n if n is even 2 nC will be maximum r When r=n2–1 or n+1 if n is odd 2 The term containing greatest binomial coefficient will be middle term in the expansion of (1 + x)n E3
JEE-Mathematics Illustration 7 : x3 9 Find the middle term in the expansion of 3x 6 Solution : x3 9 The number of terms in the expansion of 3x 6 is 10 (even). So there are two middle terms. i.e. 9 1 th and 9 3 th are two middle terms. They are given by T5 and T6 2 2 x3 4 x12 9.8.7.6 35 189 x17 6 64 1.2.3.4 . 8 T5 = T4 +1 = 9 C 4( 3 x ) 5 = 9C435x5. = 24.34 x17 = x3 5 x15 9.8.7.6 . 34 – 21 x19 6 – 9C434.x4. 65 1.2.3.4 25.35 16 and T6 = T5+1 = 9C5(3x)4 = = x19 = Ans. ( c ) Term independent of x : Term independent of x does not contain x ; Hence find the value of r for which the exponent of x is zero. Illustration 8 : x 3 10 Solution : The term independent of x in 3 is - 2 x 2 (A) 1 5 (C) 10 C1 (D) none of these (B) 12 General term in the expansion is r 10r 3r 35r 2 10 Cr x 2 3 2 10 C x 10 . 10r For constant term, 3r 10 r 20 3 2x2 23 r 22 which is not an integer. Therefore, there will be no constant term. Ans. (D) Do yourself - 2 : (i) Find the 7th term of 3 x 2 1 10 3 (ii) Find the term independent of x in the expansion : 2 x 2 3 25 x3 (iii) Find the middle term in the expansion of : (a) 2x 3 6 (b) 2 x 2 1 7 3 2x x (d) Numerically greatest term : NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#11\\Eng\\01.BINOMIAL.p65 Let numerically greatest term in the expansion of (a + b)n be T . r+1 | Tr 1 | Tr where T = nC an–rbr Tr 1 Tr 2 r+1 r Solving above inequalities we get n 1 1 r n 1 1 a 1 a bb n 1 Case I : When 1 a is an integer equal to m, then T and T will be numerically greatest term. m m+1 b n 1 Case II : When 1 a is not an integer and its integral part is m, then T will be the numerically m+1 b greatest term. 4E
JEE-Mathematics Illustration 9 : Find numerically greatest term in the expansion of (3 – 5x)11 when x = 1 5 n 1 1 r n 1 Using 1 a 1 a Solution : bb 11 1 1 r 11 1 1 3 1 3 5 x 5 x solving we get 2 < r < 3 r = 2, 3 so, the greatest terms are T2+1 and T3+1. Greatest term (when r = 2) T3 = 11C2.39 (–5x)2 = 55.39 = T4 Ans. From above we say that the value of both greatest terms are equal. Illustration 10 : Given T in the expansion of (1 – 3x)6 has maximum numerical value. Find the range of 'x'. 3 Solution : Using n 1 1 r n 1 1 a 1 a bb 6 1 1 2 7 1 1 1 1 3x 3x Let |x| = t 21t 1 2 21t 3t 1 3t 1 21t 3 4t 1 0 t 1 , 1 3t 1 3t 1 3 4 21t 1 2 15t 2 3 15 2 3t 1 0 t , , 3t 1 Common solution t 2 , 1 x 1 , 2 2 , 1 15 4 4 15 1 5 4 Do yourself -3 : ( i ) Find the numerically greatest term in the expansion of (3 – 2x)9, when x = 1. NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#11\\Eng\\01.BINOMIAL.p65 (ii) In the expansion of 1 2x n when 1 2 3 x = – , it is known that 3rd term is the greatest term. Find the 2 possible integral values of n. 4. PROPERTIES OF BINOMIAL COEFFICIENTS : E n (1+x)n nCrrr ; n N = C + Cx + C x2 + C x3 +.........+C xn ....(i) 0 1 2 3 n ....(ii) r 0 ...(iii) where C ,C ,C ,............C are called combinatorial (binomial) coefficients. 012 n ( a ) The sum of all the binomial coefficients is 2n. Put x = 1, in (i) we get n C + C + C + .............+ C = 2n n Cr 0 012 n r 0 ( b ) Put x=–1 in (i) we get n C – C C (1)r n C r 0 0 1 n + C –C ............+ = 0 23 r 0 5
JEE-Mathematics ( c ) The sum of the binomial coefficients at odd position is equal to the sum of the binomial coefficients at even position and each is equal to 2n–1. From (ii) & (iii), C + C + C +............ = C + C + C +....... = 2n–1 024 135 ( d ) nC + nC = n+1C r r r–1 (e) n Cr n r 1 Cn r r 1 (f) n Cr n n 1C r 1 n.n 1 n 2 C r 2 ....... n(n 1)(n 2).......(n r 1) r r r 1 r(r 1)(r 2)..........1 (g) n Cr r 1 . n 1 C n 1 r 1 Illustration 11 : Prove that : 25C + 24C +........+10C = 26C 10 10 10 11 Solution : LHS = 10C + 11C + 12C + ..............+25C 10 10 10 10 11C + 11C + 12C + .......+25C 10 11 10 10 12C + 12C +........+25C 11 10 10 13C + 13C +.........25C 11 10 10 and so on. LHS = 26C 11 Aliter : LHS = coefficient of x10 in {(1 + x)10 + (1 + x)11 +............... (1+ x)25} coefficient of x10 in x )10 {1 x}16 1 (1 1 x 1 coefficient of x10 in (1 x )26 (1 x )10 x coefficient of x11 in (1 x )26 (1 x )10 = 26C – 0 = 26C 11 11 Illustration 12 : Prove that : (i) C + 2C + 3C + ........ + nC = n . 2n–1 1 23 n (ii) C0 C1 C2 ......... Cn 2n 1 1 2 3 n 1 n 1 Solution : L.H.S. n n n n 1C r 1 r = r. nCr r 1 (i) r. . r 1 n n 1 C r 1 n 1 n 1C1 ....... n1 NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#11\\Eng\\01.BINOMIAL.p65 = n n. C 0 C n 1 r 1 = n . 2n–1 Aliter : (Using method of differentiation) (1 + x)n = nC0 + nC1x + nC2x2 + ....... + nCnxn ..........(A) Differentiating (A), we get n(1 + x)n – 1 = C + 2C x + 3C x2 + ....... + n.C xn – 1. 12 3 n Put x = 1, C1 2C2 3C3 ........ n.C n n.2n 1 n Cr 1 n n 1 nCr r 0 r 1 n 1 r0 r 1 (ii) = L.H.S. = 1 n n 1 C 1 n 1 C n 1C2 ....... n 1 C = 1 2 n 1 1 r 0 n 1 n 1 n 1 r 1 1 n 1 6 E
JEE-Mathematics Aliter : (Using method of integration) Integrating (A), we get (1 x)n 1 C C0x C1x2 C2x3 ........ C n x n 1 (where C is a constant) n 1 2 3 n 1 1 Put x = 0, we get, C = – n 1 (1 x)n1 1 C0x C1x2 C2x3 ........ C n x n 1 n 1 2 3 n 1 Put x = 1, we get C0 C1 C2 ....... Cn 2n1 1 2 3 n 1 n 1 Put x = –1, we get C0 C1 C2 ....... 1 2 3 n 1 n C12 2 2 2 (2n 1)! 2 3 n ((n 1)!)2 + x)n = n Cr xr , r 0 I l l u s t r a t i o n13:If(1 then prove that 2 .C 3.C ......... n.C Solution : (1 + x)n = C + C x + C x2 + C x3 + ........ + C xn .........(i) 01 2 2 n Differentiating both the sides, w.r.t. x, we get n(1 + x)n–1 = C + 2C x + 3C x2 + ......... + n.C xn –1 .........(ii) 12 2 n also, we have (x + 1)n = C xn + C xn – 1 + C xn –2 + ......... + C .........(iii) 01 2 n Multiplying (ii) & (iii), we get (C + 2C x + 3C x2 + ........ + C xn – 1)(C xn + C xn –1 + C xn – 2 + ......... + C ) = n(1 + x)2n – 1 12 3 n 01 2 n Equating the coefficients of xn – 1, we get C 2 2C 2 3C 2 ......... n.C 2 n . 2n 1C n 1 = (2n 1)! Ans. 1 2 3 n ((n 1)!)2 Illustration 14 : Prove that : C0 – 3C1 + 5C2 – ........(–1)n(2n + 1)Cn = 0 Solution : Tr = (–1)r(2r + 1)nCr = 2(–1)rr . nCr + (–1)r nCr n n .n1 C r 1 n (1)r nCr nn r r 0 2 (1)r .r. = 2 (1)r .n 1 C r1 (1)r . n C r r 1 r 1 r 0 Tr= NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#11\\Eng\\01.BINOMIAL.p65 = 2 n 1 C n 1 C1 ..... n C n C1 ....... = 0 0 0 Illustration 15 : Prove that (2nC0)2 – (2nC1)2 + (2nC2)2 – .... + (–1)n (2nC2n)2 = (–1)n. 2nCn ...(i) Solution : (1 – x)2n = 2nC0 – 2nC1x + 2nC2x2 – ....+(–1)n 2nC2nx2n ....(ii) and (x + 1)2n = 2nC0x2n + 2nC1x2n–1 + 2nC2x2n–2 +...+2nC2n Multiplying (i) and (ii), we get .... + 2nC2n) (x2 –1)2n = (2nC0 – 2nC1x + .... + (–1)n 2nC2nx2n) × (2nC0x2n + 2nC x2n–1 + .... (iii) 1 Now, coefficient of x2n in R.H.S. = (2nC0)2 – (2nC1)2 + (2nC2)2 – ...... + (–1)n (2nC2n)2 General term in L.H.S., T = 2nC (x2)2n – r(–1)r r+1 r Putting 2(2n – r) = 2n r=n T = 2nC x2n(–1)n n+1 n Hence coeffiecient of x2n in L.H.S. = (–1)n.2nCn But (iii) is an identity, therefore coefficient of x2n in R.H.S. = coefficient of x2n in L.H.S. (2nC0)2 – (2nC1)2 + (2nC2)2 – .... + (–1)n (2nC2n)2 = (–1)n. 2nCn E7
JEE-Mathematics Il lu str ati on 16 : Prove that : nC0.2nCn – nC1. C2n–2 nn + nC2.2n–4Cnn + .... = 2n Solution : L.H.S. = Coefficient of xn in [nC0(1 + x)2n – nC1(1 + x)2n – 2 ......] = Coefficient of xn in [(1 + x)2 – 1]n = Coefficient of xn in xn(x + 2)n = 2n Illustration 17 : If (1 + x)n = C0 + C1 x + C2x2 + ..... + Cnxn then show that the sum of the products of the Ci's taken two at a time represented by : CiCj is equal to 22n – 1 – 2n! Since (C0 + C1 + C2 +.....+ Cn–1 + Cn)2 2.n !n ! 0 i j n Solution : = C 2 C12 C 2 ..... C 2 1 C 2 2(C0C1 C0C2 C0C3 ... C0Cn + C1C2 + C1C3 +... 0 2 n n + C1Cn+ C2C3 + C2C4+...+C2Cn +.....+Cn–1Cn) (2n)2 = 2nCn + 2 CiC j 0 i jn Hence CiCj 2 2 n 1 2n! Ans. 2.n !n ! 0i jn Illu stration 18 : If (1 + x)n = C0 + C1x + C2 x2 +.....+ Cnxn then prove that C i C j 2 = (n – 1)2nCn + 22n 0i jn Solution : L.H.S C i C j 2 0i jn = (C0 + C1)2 + (C0 + C2)2 +....+ (C0 + Cn)2 + (C1 + C2)2 + (C1 + C3)2 +....+ (C1 + Cn)2 + (C2 + C3)2 + (C2 + C4)2 +... + (C2 + Cn)2 +....+ (Cn – 1 + Cn)2 = n (C 2 C 2 C 2 .... C 2 ) 2 CiCj 0 1 2 n 0i jn = n.2nCn + 2. 22n1 2n ! ! {from Illustration 17} 2.n ! n = n .2nCn + 22n – 2nCn = (n – 1) . 2nCn + 22n = R.H.S. Do yourself - 4 : (i) n n n ........ n = 0 1 2 n (A) 2n – 1 (B) 2nC (C) 2n (D) 2n+1 n (ii) If (1 + x)n = C + C x + C x2 + .......... + C xn, n N. Prove that 01 2 n (a) 3C – 8C + 13C – 18C + .......... upto (n + 1) terms = 0, if n 2. 0 1 2 3 (b) 2C + 22 C1 23 C2 24 C3 ...... 2n 1 C n 3n1 1 NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#11\\Eng\\01.BINOMIAL.p65 02 3 4 n 1 n 1 (c) C 2 C12 C 2 ...... C 2 (2n 1)! 0 2 2 n ((n 1)!)2 3 n 1 5 . MULTINOMIAL THEOREM : n Using binomial theorem, we have (x + a)n = n Cr xnr ar , n N r 0 n n ! xnrar n! xsar r)!r! r!s! r0 (n = , where s r = n = + rs n This result can be generalized in the following form. (x1 + x2 + ...... + xk)n = r1 ! r2 n! ! x1r1 x r2 .... ..x rk !.....rk 2 k r1 r2 .....rk n 8 E
JEE-Mathematics The general term in the above expansion r1 ! r2 n ! ! . x1r1 x r2 x r3 ......x rk ! r3 !.....rk 2 3 k The number of terms in the above expansion is equal to the number of non-negative integral solution of the equation r1 + r2 + ....... + rk = n because each solution of this equation gives a term in the above expansion. The number of such solutions is nk1 C k1 Particular cases : (i) + + z)n n! xryszt (x y = r!s!t! r s t n The above expansion has Cn + 3 –1 = Cn + 2 terms 3–1 2 n! xpyqzrus pqrsn p !q !r !s! (ii) (x + y + z + u)n = There are C =n+4–1 Cn+3 terms in the above expansion. 4–1 3 Illu stration 19 : Find the coefficient of x2 y3z4w in the expansion of (x – y – z + w)10 Solution : (x – y – z + w)10 = n! (x)p (y)q (z)r (w)s pq r s10 p ! q ! r ! s ! We want to get x2y3z4w this implies that p = 2, q = 3, r = 4, s = 1 10! Ans. Coefficient of x2y3z4w is 2! 3! 4! 1! (–1)3(–1)4 = –12600 Ans. Illustration 20 : Find the total number of terms in the expansion of (1 + x + y)10 and coefficient of x2y3. Solution : Total number of terms = C10+3–1 = 12C = 66 3–1 2 10! Coefficient of x2y3 = 2! 3! 5! 2520 Illustration 21 : Find the coefficient of x5 in the expansion of (2 – x + 3x2)6. Solution : The general term in the expansion of (2 – x + 3x2)6 = 6 ! 2r (x)s (3x2 )t , where r + s + t = 6. r!s!t! NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#11\\Eng\\01.BINOMIAL.p65 = 6 ! 2r (1)s (3)t xs2t r!s!t! For the coefficient of x5, we must have s + 2t = 5. But, r + s + t = 6, s = 5 – 2t and r = 1 + t, where 0 r, s, t 6. Now t = 0 r = 1, s = 5. t = 1 r = 2, s = 3. t = 2 r = 3, s = 1. Thus, there are three terms containing x5 and coefficient of x5 = 6 ! 21 (1)5 30 6 ! 22 (1)3 31 6 ! 23 (1)1 32 1! 5! 0! 2! 3! 1! 3! 1! 2! = –12 – 720 – 4320 = –5052. Ans. E9
JEE-Mathematics 2n (a) ar = a2n–r (b)n 1 1 (3n an ) 2 Illustration 22 : If (1+x+x2)n ar xr , then prove that ar r 0 r 0 Solution : (a) We have n 2n arxr 1 x x2 ....(A) r 0 1 Replace x by x 1 1 1 n 2n 1 r x x2 x ar r 0 x2 x 1 n 2n ar x2nr r 0 2n 2n {Using (A)} ar xr ar x2nr r0 r0 Equating the coefficient of x2n–r on both sides, we get a2n–r = ar for 0 < r < 2n. Hence ar = a2n–r. (b) Putting x=1 in given series, then a0 + a1 + a2 + .........+ a2n = (1+1+1)n ....(1) a0 + a1 + a2 + ..........+ a2n = 3n But ar = a2n–r for 0 < r < 2n series (1) reduces to 2(a0 + a1 +a2 + ........+ an–1) + an = 3n. a0 + a1 +a2 + .......+ an–1 = 1 (3n – an) 2 Do yourself - 5 : ( i ) Find the coefficient of x2y5 in the expansion of (3 + 2x – y)10. 6 . APPLICATION OF BINOMIAL THEOREM : Illustration 23 : If 6 2 n 1 = [N] + F and F = N – [N]; where [.] denotes greatest integer function, then 6 14 NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#11\\Eng\\01.BINOMIAL.p65 NF is equal to (B) an even integer (C) odd integer (D) 402n+1 (A) 202n+1 Solution : 2n1 Since 6 6 14 = [N] + F Let us assume that f = 6 6 14 2n1 ; where 0 f < 1. 2n1 2n 1 Now, [N] + F – f = 6 6 14 – 6 6 14 6 2n (14 ) 2n1 C 3 6 6 2n2 (14 )3 .... =22 n 1 6 C1 [N] + F – f = even integer. Now 0 < F < 1 and 0 < f < 1 so –1 < F – f < 1 and F – f is an integer so it can only be zero 2 n1 2n 1 6 14 6 14 Thus NF =6 6 = 202n+1. Ans. (A,B) 10 E
JEE-Mathematics Illustration 24 : Find the last three digits in 1150. Solution : Expansion of (10 + 1)50 = 50C 1050 + 50C 1049 + ..... +50C 102 + 50C 10 + 50C 01 48 49 50 = 50 C0105050C11049...... 50C47103 + 49 × 25 × 100 + 500 + 1 1000K 1000 K + 123001 Last 3 digits are 001. Illustration 25 : Prove that 22225555 + 55552222 is divisible by 7. Solution : When 2222 is divided by 7 it leaves a remainder 3. So adding & subtracting 35555, we get : E 22225555 35555 3555555552222 E1 E2 For E : Now since 2222–3 = 2219 is divisible by 7, therefore E is divisible by 7 11 ( xn – an is divisible by x –a) For E2 : 5555 when devided by 7 leaves remainder 4. So adding and subtracting 42222, we get : E = 35555 + 42222 + 55552222 – 42222 2 = (243)1111 + (16)1111 + (5555)2222 – 42222 Again (243)1111 + 161111 and (5555)2222 – 42222 are divisible by 7 ( xn + an is divisible by x + a when n is odd) Hence 22225555 + 55552222 is divisible by 7. Do yourself - 6 : ( i ) Prove that 525 – 325 is divisible by 2. ( i i ) Find the remainder when the number 9100 is divided by 8. (i i i ) Find last three digits in 19100. ( i v ) Let R (8 3 7 )20 and [.] denotes greatest integer function, then prove that : 1 (a) [R] is odd (b) R [R] 1 (8 3 7 )20 ( v ) Find the digit at unit's place in the number 171995 + 111995 – 71995. 7. BINOMIAL THEOREM FOR NEGATIVE OR FR ACTIONAL INDICES : E If n Q, then (1 + x)n = 1 + nx + n(n 1) x2 + n(n 1)(n 2) x3 + ....... provided | x | < 1. NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#11\\Eng\\01.BINOMIAL.p65 2! 3! Note : (i) When the index n is a positive integer the number of terms in the expansion of ( 1+ x)n is finite i.e. (n+1) & the coefficient of successive terms are : nC , nC , nC , ....... nC n 012 (ii) When the index is other than a positive integer such as negative integer or fraction, the number of terms in the expansion of (1+ x)n is infinite and the symbol nC cannot be used to denote the coefficient of the r general term. (iii) Following expansion should be remembered (|x| < 1). (a) (1 + x)-1 =1 – x + x2 – x3 + x4 - .... (b) (1 – x)–1 =1 + x + x2 + x3 + x4 + .... (c) (1 + x)-2 =1 – 2x + 3x2 – 4x3 + .... (d) (1 – x)–2 =1 + 2x + 3x2 + 4x3 + .... (e) (1 + x)–3 = 1 – 3x + 6x2 – 10x3 + ..... + (1)r (r 1)(r 2) xr ........ 2! 11
JEE-Mathematics (f) (1 – x)–3 = 1 + 3x + 6x2 + 10x3 + ..... + (r 1)(r 2) xr ........ 2! (iv) The expansions in ascending powers of x are only valid if x is ‘small’. If x is large i.e. | x |>1 then we may find it convenient to expand in powers of 1/x, which then will be small. 8 . APPROXIMATIONS : (1 + x)n = 1 + nx n(n 1) n(n 1)(n 2) x3 ....... + x2 + 1.2 1.2.3 If x < 1, the terms of the above expansion go on decreasing and if x be very small, a stage may be reached when we may neglect the terms containing higher powers of x in the expansion. Thus, if x be so small that its square and higher powers may be neglected then (1 + x)n = 1 + nx, approximately. This is an approximate value of (1 + x)n Illustration 26 : If x is so small such that its square and higher powers may be neglected then find the approximate value of (1 3x )1 / 2 (1 x )5 / 3 4 x 1 / 2 (1 3x )1 / 2 (1 x )5 / 3 1 3 x 1 5x 1 19 x 1 x 1 /2 1 19 x 1 x 23 2 6 4 2 6 8 Solution : 4 x 1 / 2 = = 2 = 2 1/2 2 1 x 4 = 1 2 x 19 x = 1 x 19 x =1– 41 x Ans. 2 4 6 8 12 24 Illustration 27 : The value of cube root of 1001 upto five decimal places is – (A) 10.03333 (B) 10.00333 (C) 10.00033 (D) none of these 1 1 1 / 3 1 1 1 1 / 3(1 / 3 1) 1 ..... 1000 3 . 2! 10002 Solution : (1001)1/3 = (1000+1)1/3 = 10 = 10 1000 = 10{1 + 0.0003333 – 0.00000011 + .....} = 10.00333 Ans. (B) Illustration 28 : The sum of 1 + 1 1.3 1.3.5 .... is - 4 4.8 4.8.12 (A) 2 1 (C) 3 (D) 23/2 (B) 2 Solution : Comparing with 1 + nx + n(n 1) x2 .... NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#11\\Eng\\01.BINOMIAL.p65 2! ....... (i) nx = 1/4 (by (i)) and n(n 1)x2 = 1.3 2! 4.8 or nx(nx x) 3 1 1 x 3 = 32 4 4 16 2! 1 x 3 x 1 3 1 ........(ii) 4 4 4 4 2 putting the value of x in (i) n (–1/2) = 1/4 n = –1/2 sum of series = (1 + x)n = (1 – 1/2)–1/2 = (1/2)–1/2 = 2 Ans. (A) 12 E
JEE-Mathematics 9. EXPONENTIAL SERIES : ( a ) e is an irrational number lying between 2.7 & 2.8. Its value correct upto 10 places of decimal is 2.7182818284. ( b ) Logarithms to the base ‘e’ are known as the Napierian system, so named after Napier, their inventor. They are also called Natural Logarithm. x x2 x3 1 1 n 1 ....... ; where x may be any real or complex number & e = n (c) ex = Lim 1! 2! 3! n ( d ) ax = 1 x na x2 n2a x3 n3a ....... , where a > 0 1! 2! 3! 111 ( e ) e = 1 ....... 1! 2! 3! 10. LOGARITHMIC SERIES : (a) n (1 + x) =x x2 x3 x4 ....... , where –1 < x 1 234 (b) n (1 - x) = - x x2 x3 x4 ....... , where –1 x1 234 Remember : (i) 1 1 1 1 ....... n 2 (ii) elnx = x ; for all x > 0 234 (iv) n10 = 2.303 (iii) n2 = 0.693 NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#11\\Eng\\01.BINOMIAL.p65 ANSWERS FOR DO YOURSELF 1. (i) 5C x (3x2)5 + 5C (3x2)4 x + 5C (3x2)3 x 2 + 5C (3x2)2 x 3 + 5C (3x2)1 x 4 + 5C x 5 0 1 2 2 2 3 2 4 2 5 2 2: 3. (ii) nC yn + nC yn–1.x + nC .yn–2.x2 + ..........+nC .xn 4. 01 2 n 5. 6. (i) 70 x8 ; (ii) 1 25 ! ! 215 310 ; (i ii ) (a) –20; (b) –560x5, 280x2 3 0! 5 E (i) 4th & 5th i.e. 489888 ( i i ) n = 4, 5, 6 (i) C (i) –272160 or – 10C × 5C × 108 52 (ii) 1 (iii) 801 (v) 1 13
JEE-Mathematics CHECK YOUR GRASP EXERCISE - 01 SELECT THE CORRECT ALTERNATIVE (ONLY ONE CORRECT ANSWER) 1. If the coefficients of x7 & x8 in the expansion of 2 x n are equal , then the value of n is - 3 (A) 15 (B) 45 (C) 55 (D) 56 2. The sum of the binomial coefficients of 2 x 1 n is equal to 256 . The constant term in the expansion is - x (A) 1120 (B) 2110 (C) 1210 (D) none 3 . The sum of the coefficients in the expansion of (1 2x + 5x2)n is ' a ' and the sum of the coefficients in the expansion of (1 + x)2n is b . Then - (A) a = b (B) a = b2 (C) a2 = b (D) ab = 1 4 . Given that the term of the expansion (x1/3 x1/2)15 which does not contain x is 5 m where m N , then m is equal to - (A) 1100 (B) 1010 (C) 1001 (D) none 5 . The expression 1 1 7 4 x 1 7 is a polynomial in x of degree - 4x 1 1 4 x 1 2 2 (A) 7 (B) 5 (C) 4 (D) 3 6 . In the binomial (21/3 + 31/3)n, if the ratio of the seventh term from the beginning of the expansion to the seventh term from its end is 1/6 , then n is equal to - (A) 6 (B) 9 (C) 12 (D) 15 7. The term independent of x in the product (4 x 7 x2 ) x 3 11 is - x (A) 7.11C6 (B) 36. 11C6 (C) 35. 11C5 (D) –12. 211 8 . If ‘a’ be the sum of the odd terms & ‘b’ be the sum of the even terms in the expansion of (1 + x)n , then (1 x²)n is equal to - (A) a² b² (B) a² + b² (C) b² a² (D) none 9 . The sum of the coefficients of all the even powers of x in the expansion of (2x2 3x + 1)11 is - (A) 2 . 610 (B) 3 . 610 (C) 611 (D) none 1 0 . The greatest terms of the expansion (2x + 5y)13 when x = 10, y = 2 is - NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#11\\Eng\\01.BINOMIAL.p65 (A) 13C . 208 . 105 (B) 13C . 207 . 104 (C) 13C . 209 . 104 (D) none of these 5 6 4 1 1 . Number of rational terms in the expansion of 2 4 3 100 is - (A) 25 (B) 26 (C) 27 (D) 28 p n 1 2 .If q 0 for p < q, where p, q W, then 2r r 0 (A) 2n (B) 2n–1 (C) 22n–1 (D) 2nCn 47 5 52 j x x 1 3 . 4 y , then = j1 3 y (A) 11 (B) 12 (C) 13 (D) 14 14 E
JEE-Mathematics 11 1 1 14. If nN& n is even , then ...... = 1. (n 1) ! 3 ! (n 3) ! 5 ! (n 5) ! (n 1) ! 1 ! (A) 2n 2n 1 (C) 2n n ! (D) none of these (B) n ! 1 5 . Let R (5 5 11)31 I ƒ , where I is an integer and ƒ is the fractional part of R, then R · ƒ is equal to - (A) 231 (B) 331 (C) 262 (D) 1 10 10 15 1 6 . The value of r 14 r is equal to - r0 (A) 25C12 (B) 25C15 (C) 25C10 (D) 25C11 1 7 . C0 C1 C2 ...... C10 is equal to (here C = 10C ) 311 1 123 11 rr (D) 211 211 1 311 11 (A) (B) (C) [JEE 98] 11 11 11 (D) none of these (D) 01 n 1 nr 1 8 . If a n r 0 n C r , then r0 n C r equals - (A) ( n -1) a (B) n a (C) n a /2 n n n 1 9 . The last two digits of the number 3400 are - (A) 81 (B) 43 (C) 29 SELECT THE CORRECT ALTERNATIVES (ONE OR MORE THAN ONE CORRECT ANSWERS) 2 0 . If the coefficients of three consecutive terms in the expansion of (1 + x)n are in the ratio of 1 : 7 : 42, then n is divisible by - (A) 9 (B) 5 (C) 3 (D) 11 3 1 20 46 21. In the expansion of 4 - (A) the number of irrational terms = 19 (B) middle term is irrational (C) the number of rational terms = 2 (D) 9th term is rational NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#11\\Eng\\01.BINOMIAL.p65 2 2 . If (1 + x + x2 + x3)100 = a0 + a1x + a2x2 + ......... + a300x300, then - (A) a0 + a1 + a2 + a3 +.......+ a300 is divisible by 1024 (B) a0 + a2 + a4 +.......+ a300 = a1 + a3 + ....... + a299 (C) coefficients equidistant from beginning and end are equal (D) a1 = 100 2 3 . The number 101100 1 is divisible by - (A) 100 (B) 1000 (C) 10000 (D) 100000 n 2 4 . If 9 80 = I + f where I , n are integers and 0 < f < 1 , then - (A) I is an odd integer (B) I is an even integer (C) (I + f) (1 f) = 1 n (D) 1 f = 9 80 1 30 x 25. In the expansion of x 2 / 3 , a term containing the power x13 - (A) does not exist (B) exists and the co-efficient is divisible by 29 (C) exists and the co-efficient is divisible by 63 (D) exists and the co-efficient is divisible by 65 E 15
JEE-Mathematics 2 6 . The co-efficient of the middle term in the expansion of (1 + x)2n is - 1.3.5.7......(2 n 1) (B) 2nC (A) n ! 2n n (C) (n 1) (n 2) (n 3) .... (2n 1) (2n) 2 .6 .10 .14 ...... (4n 6) (4n 2) 1.2.3.......... (n 1) n (D) 1.2.3 .4 .....(n 1) . n CHECK YOUR GRASP ANSWER KEY EXERCISE-1 NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#11\\Eng\\01.BINOMIAL.p65 Que. 1 2 3 4 5 6 7 8 9 10 Ans. C A A C D B B A B C Que. 11 12 13 14 15 16 17 18 19 20 Ans. B B C B C D B C D B,D Que. 21 22 23 24 25 26 Ans. A,B,C,D A,B,C,D A,B,C A,C,D B,C,D A,B,C,D 16 E
EXERCISE - 02 JEE-Mathematics BRAIN TEASERS SELECT THE CORRECT ALTERNATIVES (ONE OR MORE THAN ONE CORRECT ANSWERS) 1 . The coefficient of xr (0 r n 1) in the expression : (x + 2)n1 + (x + 2)n2 . (x + 1) + (x + 2)n3 . (x + 1)² + ...... + (x + 1)n1 is - (A) nC (2r 1) (B) nC (2nr 1) (C) nC (2r + 1) (D) nC (2nr + 1) r r r r 2 . If (1 + x + x²)25 = a + a x + a x² + ..... + a . x50 then a + a + a + ..... + a is - 01 2 50 024 50 (A) even (B) odd & of the form 3n (C) odd & of the form (3n 1) (D) odd & of the form (3n + 1) 3 . The co-efficient of x4 in the expansion of (1 x + 2x2)12 is - (A) 12C (B) 13C (C) 14C (D) 12C + 3 13C + 14C 3 3 4 3 34 4 . Let (1 + x2)2 (1 + x)n = A + A x + A x2 + ...... If A , A , A are in A.P. then the value of n is - 01 2 012 (A) 2 (B) 3 (C) 5 (D) 7 nr 5 . If n kC = xC then - ry k 1 (A) x = n + 1 ; y = r (B) x = n ; y = r + 1 (C) x = n ; y = r (D) x = n + 1 ; y = r + 1 6 . Co-efficient of t in the expansion of ( + p)m 1 + ( + p)m 2 ( + q) + ( + p)m 3 ( + q)2 +...... ( + q)m 1 where q and p q is - m C t pt q t m C t pm t q m t m C t pt q t m C t pm t q m t (A) (B) (C) (D) pq pq pq pq 7 . The co-efficient of x401 in the expansion of (1 + x + x2 + ...... + x9) 1 , (x < 1) is - (A) 1 (B) 1 (C) 2 (D) 2 8 . Number of terms free from radical sign in the expansion of (1 + 31/3 + 71/7)10 is - (A) 4 (B) 5 (C) 6 (D) 8 9. 30 15 30 15 30 15 The value r for which r r r 1 1 ....... 0 r is maximum is/are - (A) 21 (B) 22 (C) 23 (D) 24 NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#11\\Eng\\01.BINOMIAL.p65 10. If the 6th term in the expansion of 3 xn when x = 3 is numerically greatest then the possible integral 2 3 value(s) of n can be - (A) 11 (B) 12 (C) 13 (D) 14 1 1 . In the expansion of (1 + x)n (1 + y)n (1 + z)n , the sum of the co-efficients of the terms of degree ' r ' is - (A) n3 C r (B) n C r3 (C) 3nC (D) 3 . 2nC r r 1 2 . 35 10 4 5 r x 6 r 0 5 y , then x – y is equal to - (A) 39 (B) 29 (C) 52 (D) 40 sn n CssCr is - 1 3 . The value of r 0 s1 r s (A) 3n – 1 (B) 3n + 1 (C) 3n (D) 3(3n – 1) E 17
JEE-Mathematics 14. In the expansion of x 3 3 .2log 2 11 - x3 (A) there appears a term with the power x2 (B) there does not appear a term with the power x2 (C) there appears a term with the power x 3 1 (D) the ratio of the co-efficient of x3 to that of x 3 is 3 1 5 . The sum of the series (1² + 1).1! + (2² + 1).2! + (3² + 1).3! + ..... + (n² + 1).n! is - (A) (n + 1) . (n + 2)! (B) n . (n + 1)! (C) (n + 1) .(n + 1)! (D) none of these 1 3n x2k 16. The binomial expansion of x k , n N contains a term independent of x - (A) only if k is an integer (B) only if k is a natural number (C) only if k is rational (D) for any real k 17. Let n N. If (1 + x)n = a + ax + a x2 + ........+ a xn and a, a, a are in AP, then - 0 1 2 n n–3 n–2 n–1 (A) a , a , a are in AP (B) a , a , a are in HP 123 123 (C) n = 7 (D) n = 14 18. Set of values of r for which, 18C + 2 . 18C + 18C 20C contains - r2 r1 r 13 (A) 4 elements (B) 5 elements (C) 7 elements (D) 10 elements BRAIN TEASERS ANSWER KEY EXERCISE-2 NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#11\\Eng\\01.BINOMIAL.p65 Que. 1 2 3 4 5 6 7 8 9 10 Ans. B A Que. 11 12 D A,B B B B C B,C B,C,D Ans. C D 13 14 15 16 17 18 A B,C,D B D A,C C 18 E
JEE-Mathematics EXERCISE - 03 MISCELLANEOUS TYPE QUESTIONS FILL IN THE BLANKS 1 . The greatest binomial coefficient in the expansion of (a + b)n is ________ given that the sum of all the coefficients is equal to 4096. 2 . The number 71995 when divided by 100 leaves the remainder ________. 3. The term independent of x in the expansion of x 2 1 15 is ________ . x 4 . If (1 + x + x² + ..... + xp)n = a + a x + a x² + ..... + a xnp then a + 2 a + 3a + .... + npa = ________ . 01 2 np 1 23 np 5. If (1 + x) (1 + x + x2) (1 + x + x2 + x3) ...... (1 + x + x2 + x3 + ...... + xn) a + a x + a x2 + a x3 + ...... + a xm 01 2 3 m m then ar has the value equal to ________ . r 0 6. If the 6th term in the expansion of the binomial 1 x2 log10 x 8 is 5600, then x = ________ . x 8 / 3 7 . (1 + x) (1 + x + x2) (1 + x + x2 + x3) ...... (1 + x + x2 + ...... + x100) when written in the ascending power of x then the highest exponent of x is ________ . MATCH THE COLUMN Following question contains statements given in two columns, which have to be matched. The statements in Column-I are labelled as A, B, C and D while the statements in Column-II are labelled as p, q, r and s. Any given statement in Column-I can have correct matching with ONE statement in Column-II. 1 . Column-I Column-II (A) (2n + 1) (2n + 3) (2n + 5) ....... (4n 1) is equal to (n 1)n (p) n! (B) C1 2 . C2 3 . C 3 ...... n . C n is equal to (q) n . 2n . (2n 1) C0 C1 C2 Cn 1 (4n) ! n ! here C stand for nC . (r) 2n . (2n) ! (2n) ! rr n (n 1) (C) If (C0 + C1) (C1 + C2) (C2 + C3) ...... (Cn1 + Cn) (s) = m . C C C .... C , then m is equal to 2 123 n1 (D) If C are the binomial coefficients in the expansion of r nn NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#11\\Eng\\01.BINOMIAL.p65 (1 + x)n, the value of (i j) C C is i j i 1j1 ASSERTION & REASON These questions contains, Statement I (assertion) and Statement II (reason). (A) Statement-I is true, Statement-II is true ; Statement-II is correct explanation for Statement-I. (B) Statement-I is true, Statement-II is true ; Statement-II is NOT a correct explanation for statement-I (C) Statement-I is true, Statement-II is false (D) Statement-I is false, Statement-II is true 1 . Statement-I : Coefficient of ab8c3d2 in the expansion of (a + b + c + d)14 is 180180 Because Statement-II : General term in the expansion of (a1 + a2 + a3 + ..... + am)n = n1 !n2 n! ! a1n1 a2n2 ... anmm , where n1 + n2 + n3 + ... + nm = n. !n3 !....nm (A) A (B) B (C) C (D) D E 19
JEE-Mathematics 2 . 1 15 15Crpr q15r 1 Statement-I : If q = 3 and p + q = 1, then 15 5 r 3 r 0 Because n Statement-II : If p + q = 1 , 0 < p < 1, then r nC rpr qnr np r 0 (A) A (B) B (C) C (D) D 3 . Statement-I : The greatest value of 40C0 . 60Cr + 40C1 . 60Cr–1........40C40 . 60Cr–40 is 100C50 Because Statement-II : The greatest value of 2nCr, (where r is constant) occurs at r = n. (A) A (B) B (C) C (D) D 4. Statement-I : If x = nCn–1 + n+1Cn–1 + n+2Cn–1 + .......... + 2nCn–1, then x 1 is integer. 2n 1 Because Statement-II : nCr + nCr–1 = n+1Cr and nCr is divisible by n if n and r are co-prime. (A) A (B) B (C) C (D) D COMPREHENSION BASED QUESTIONS Comprehension # 1 2n If n is positive integer and if (1 + 4x + 4x2)n = ar xr , where ai's are (i = 0, 1, 2, 3, ..... , 2n) real r 0 numbers. On the basis of above information, answer the following questions : n 1 . The value of 2 a2r is - r 0 (A) 9n – 1 (B) 9n + 1 (C) 9n – 2 (D) 9n + 2 n 2 . The value of 2 a2r 1 is - r 1 (A) 9n – 1 (B) 9n + 1 (C) 9n – 2 (D) 9n + 2 (C) n.22n (D) (n + 1).22n 3 . The value of a2n–1 is - (C) 8n2 – 4n (D) 8n – 4 (A) 22n (B) (n – 1).22n 4 . The value of a2 is - NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#11\\Eng\\01.BINOMIAL.p65 (A) 8n (B) 8n2 – 4 MISCELLANEOUS TYPE QUESTION ANSWER KEY EXERCISE-3 Fill in the Blanks 7. 5050 1 . 12C6 2 . 43 3 . 3003 np 6 . x = 10 E Match the Column 4 . (p + 1)n 5 . (n + 1) ! 2 1 . (A)(r), (B)(s), (C)(p), (D)(q) Assertion & Reason 1. C 2. D 3. C 4. A Comprehension Based Questions Comprehension # 1 : 1. B 2. A 3. C 4. C 20
EXERCISE - 04 [A] JEE-Mathematics CONCEPTUAL SUBJECTIVE EXERCISE 1 . If the coefficients of (2r + 4)th, (r - 2)th terms in the expansion of (1 + x)18 are equal, find r. 2 . If the coefficients of the rth, (r +1)th & (r + 2 )th terms in the expansion of (1 + x)14 are in AP, find r. 3. Find the term independent of x in the expansion of : (a) x 3 10 (b) 1 x1/ 3 x 1 / 5 8 2 3 2x2 4 . Prove that : n -1C + n -2C + n -3C +....... + rC = n C . r rr r r+1 5. If 40C . x(1 x)39 + 2 . 40C x2 (1 x)38 + 3 40C x3 (1 x)37 + ...... + 40. 40C x40 = ax + b, then find a & b. 1 2 3 40 6 . If n + 1C + 2 (2C + 3C + 4C + ...... + nC ) = 12 + 22 + 32 + ......... + 1002, then find n. 2 222 2 7 . Which is larger : (9950 +10050) or (101)50. C2n -2 + 4n 8. Show that 2n -2C + 2. n-1 C2n -2 > , n N, n >2 n-2 n n 1 9 . Find the coefficient of x4 in the expansion of : (a) (1 + x + x2 + x3)11 (b) (2 – x + 3x2)6 1 0 . Find numerically the greatest term in the expansion of : 3 1 (a) ( 2 + 3x)9 when x = (b) ( 3 – 5x)15 when x = 2 5 11. Prove that the ratio of the coefficient of x10 in (1– x2)10 & the term independent of x in x 2 10 is 1: 32. x 3x2 9 1 1 2 . Find the term independent of x in the expansion of (1+ x +2x3) 2 3x . n 1 3 . Prove that n Ck sin Kx. cos ( n - K)x = 2n -1 sin nx. k 0 1 4 . Find the coefficient of : NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#11\\Eng\\01.BINOMIAL.p65 (a) x6 in the expansion of (ax2 + bx + c)9. (b) x2 y3 z4 in the expansion of (ax - by + cz)9. (c) a2 b3 c4d in the expansion of (a – b – c + d)10. 20 20 30 xCy Prove 25 30 70 100 C25 1 5 . (a) , then find x, y. (b) that : r 0 r 25 r 25 r r r0 n r n n k 16. Prove that : r k k r k 1 7 . 25 1 r 30 30 Prove that : r 0 2 5 0 r r n2 n 1 n 2n 1 18. Prove that : r r 2 n 2 r0 E 21
JEE-Mathematics Prove the following (here C = nC ) (Q. 19 to 26) : rr 1 9 . C C + C C + C C +.......+ C C = (2n)! 01 12 23 n-1 n (n 1)!(n 1)! 2n! 2 0 . C0 Cr C1Cr 1 C2C r 2 ....... C nrC n (n r )!(n r )! 2 1 . C 2 + C 2 + C 2 +....... + C 2 = (2n)! 012 n n!n! 22. C0 C1 C2 C3 ....... (1)r .C r (1)r (n 1)! r!.(n r 1)! 2 3 . C + 2C + 3C + ....... + n. C =n. 2n-1 1 23 n 2 4 . C + 2C + 3C +....... + (n +1) C = (n + 2) 2n-1 0 12 n 25. C0 C1 C2 ....... Cn 2n1 1 2 3 n 1 n 1 26. C1 2C2 3C3 ...... n.Cn n(n 1) C0 C1 C2 C n 1 2 27. Prove the identity 1 1 2n 2 1 . 2n 1 C r C2 n 1 2n 1 2n Cr r 1 2 8 . If (1 + x )15 = C + C . x + C . x2 +....+ C . x15 and C + 2C + 3C +....+ 14C = a2b + c, then find a + b + c. 01 2 15 23 4 15 215 30 30 214 30 29 213 30 28 ...... 30 15 0 15 1 14 2 13 15 29. Evaluate : 0 CONCEPTUAL SUBJECTIVE EXERCISE ANSWER KEY EXERCISE-4(A) NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#11\\Eng\\01.BINOMIAL.p65 1. r = 6 2 . r = 5 or 9 3. (a) T = 5 (b) T = 7 5. a = 40, b = 0 6. 100 3 12 6 7 . 10150 7.313 17 9 . (a) 990 (b) 3660 1 0 . (a) T7 2 (b) 455 x 312 12. 54 1 4 . (a) 84b6c3 + 630ab4c4 +756a2b2c5 + 84a3c6 ; (b) –1260.a2b3c4 ; (c) –12600 30 1 5 (a) x = 50, y = 25 2 8 . 2 8 29. 15 22 E
EXERCISE - 04 [B] JEE-Mathematics BRAIN STORMING SUBJECTIVE EXERCISE 2n 2n ar (x 2)r 3)r 1 .If br (x & a =1 for all k n, then show that b = C2n+1 k n n+1 r 0 r0 2 . Prove the following : (a) C02 C 2 C 2 C32 ....... (1)n C n2 0 / 2 nCn/2 if n is odd 1 2 (1)n if n is even (b) 1.C 2 + 3.C 2 + 5.C 2 +.......+(2n+1)C 2 = (n 1)(2n)! 012 n n!n! 3. (a) Find the index n of the binomial x 2 n if the 9th term of the expansion has numerically the greatest 5 5 coefficient (n N). (b) For which positive values of x is the fourth term in the expansion of (5 +3x)10 is the greatest. 4 . If a , a , a , ....... be the coefficients in the expansion of (1 + x + x2)n in ascending powers of x, then prove that : 012 (a) a a – a a + a a – .... = 0 01 12 23 (b) a a – a a + a a – ....... + a a = a or a 02 13 24 2n-2 2n n+1 n-1 (c) E1 = E2 = E3 = 3n-1 ; where E1 = a0+ a3+a6+ ....... ; E2 = a1 + a4+ a7 +....... & E3 = a2+ a5+a8+....... 5 . Prove that : 12 . C + 22 C + 32. C + 42. C . +........+(n +1)2 . C = 2n-2 ( n +1) (n + 4). 01 2 3 n If (1+ x)n = n C r .x r then prove that ; 22.C0 23.C1 24.C2 ....... 2n2.C n 3n2 2n 5 6 . 1.2 2.3 3.4 (n 1)(n 2) (n 1)(n 2) r 0 r n i n r 1 n 7. Prove that : k k 1 k 1 i0 p q p q 8 . n , p, N; p, q are constants. Prove that : j n j p q j0 9. Prove that : n n 1 n 2n 1 n r r n 1 r 1 10. Prove that : C0 C1 C2 C3 ....... Cn 1 n. 2n1 2345 n 2 (n 1)(n 2) NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#11\\Eng\\01.BINOMIAL.p65 11. Prove that : 1 n 2 n C2 3 n 4 n C4 ..... (1)n 1 n .n C n 1 2 C1 3 4 C3 5 n 1 n 1 12. Prove that : (2nC )2 +2. (2nC )2 +3.(2nC )2 +..... +2n.(2nC )2 = (4n 1)! 1 23 2n [(2n 1)!]2 BRAIN STORMING SUBJECTIVE EXERCISE ANSWER KEY EXERCISE-4(B) 23 3. (a) n = 12 (b) 5 x 20 8 21 E
JEE-Mathematics EXERCISE - 05 [A] JEE-[MAIN] : PREVIOUS YEAR QUESTIONS 1 . The sum of the coefficients in the expansion of (x + y)n is 4096. The greatest coefficient in the expansion is- [AIEEE 2002] (1) 1024 (2) 924 (3) 824 (4) 724 2 . If for positive integers r > 1, n > 2 the coefficients of the (3r)th and (r+2)th powers of x in the expansion of (1+x)2n are equal, then- [AIEEE 2002] (1) n = 2r (2) n = 3r (3) n = 2r + 1 (4) n = 2r – 1 3. If 1 x n C0 C1 x C2 x2 ... ...Cn xn , then C1 2C2 3C3 ... nCn [AIEEE-2002] C0 C1 C2 C n1 nn 1 nn 2 nn 1 n 1n 2 (1) (2) (3) (4) 2 2 2 2 4 . The number of integral terms in the expansion of 3 8 5 256 is- [AIEEE 2003] (1) 32 (2) 33 (3) 34 (4) 35 5 . The coefficient of the middle term in the binomial expansion in powers of x of (1 + x)4 and of (1 – x)6 is the same if equals- [AIEEE 2004] (1) 5 10 3 3 3 (2) 3 (3) 10 (4) 5 6 . The coefficient of xn in expansion of (1 + x)(1 – x)n is- [AIEEE 2004] (1) (n – 1) (2) (–1)n (1–n) (3) (–1)n–1 (n –1)2 (4) (–1)n–1n 7 . If the coefficients of rth, (r + 1)th and (r + 2)th terms in the binomial expansion (1 + y)m are in A.P., then m and r satisfy the equation- [AIEEE 2005] (1) m2 – m (4r – 1) + 4r2 + 2 = 0 (2) m2 – m (4r + 1) + 4r2 – 2 = 0 (3) m2 – m (4r + 1) + 4r2 + 2 = 0 (4) m2 – m(4r – 1) + 4r2 – 2 = 0 ax2 1 1 1 ax 1 11 bx bx2 8. If the coefficient of x7 in equals the coefficient of x–7 in , then a and b satisfy the relation- [AIEEE 2005] (1) ab = 1 (2) a 1 (3) a + b = 1 (4) a – b = 1 NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#11\\Eng\\01.BINOMIAL.p65 b 9 . For natural numbers m, n if (1 – y)m (1 + y)n = 1 + a1y + a2y2 +......, and a1 = a2 = 10, then (m, n) is- [AIEEE 2006] (1) (45, 35) (2) (35, 45) (3) (20, 45) (4) (35, 20) 1 0 . The sum of the series 20C0 20C1 + 20C2 20C3 + ..... ...... + 20C10 is - [AIEEE 2007] (1) 1 20C10 (2) 0 (3) – 20C10 (4) 20C10 2 1 1 . In the binomial expansion of (a b)n, n 5, the sum of 5th and 6th terms is zero, then a equals b [AIEEE 2007] 6 n 5 n4 5 (1) n 5 (2) 6 (3) 5 (4) 24 n4 E
JEE-Mathematics n [AIEEE 2008] 1 2 . Statement –1 : r 1 nCr n 2 2n1 r 0 n Statement–2 : r 1 nCr xr = (1 + x)n+nx (1+x) n–1 r 0 (1) Statement –1 is false, Statement –2 is true (2) Statement–1 is true, Statement–2 is true; Statement–2 is a correct explanation for Statement–1 (3) Statement–1 is true, Statement–2 is true; Statement–2 is not a correct explanation for Statement–1 (4) Statement–1 is true, Statement–2 is false 1 3 . The remainder left out when 82n – (62)2n+1 is divided by 9 is :- [AIEEE 2009] (1) 7 (2) 8 (3) 0 (4) 2 10 10 S3 10 j210C j . S1 j( j 1)10 C j, S2 j10 C j 1 4 . Let and [AIEEE-2010] j1 j1 j1 Statement–1 : S3 = 55 × 29. Statement–2 : S1 = 90 × 28 and S2 = 10 × 28. (1) Statement–1 is true, Statement–2 is true; Statement–2 is a correct explanation for Statement–1. (2) Statement–1 is true, Statement–2 is true; Statement–2 is not a correct explanation for statement–1. (3) Statement–1 is true, Statement–2 is false. (4) Statement–1 is false, Statement–2 is true. 1 5 . The coefficient of x7 in the expansion of (1 – x – x2 + x3)6 is :- [AIEEE 2011] (1) –144 (2) 132 (3) 144 (4) – 132 2n 2n 3 1 3 1 1 6 . If n is a positive integer, then is : [AIEEE 2012] (1) a rational number other than positive integers (2) an irrational number (3) an odd positive integer (4) an even positive integer 17. The term independent of x in expansion of x2 / 3 x 1 x 1 10 is : [JEE (Main)-2013] x1 / 3 x1 / 2 1 x NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#11\\Eng\\01.BINOMIAL.p65 (1) 4 (2) 120 (3) 210 (4) 310 PREVIOUS YEARS QUESTIONS ANSWER KEY EXERCISE-5 [A] Que. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 Ans 2 3 3 2 3 2 2 1 2 1 3 2 4 3 1 Que. 16 17 Ans 2 3 E 25
JEE-Mathematics JEE-[ADVANCED] : PREVIOUS YEAR QUESTIONS EXERCISE - 05 [B] 1. (a) For 2 r n , n + 2 n + n = [JEE 2000, (Screening ), 1+1M] r r 1 r 2 n 1 2 n 1 2 n 2 n 2 (A) r 1 r 1 r (D) r (B) (C) a ( b ) In the binomial expansion of (a - b)n, n 5, the sum of the 5th and 6th terms is zero, Then b equals - n5 n4 5 6 (A) (B) (C) (D) 6 5 n4 n5 2. For any positive integers m, n (with n m) , n = nC . Prove that : let m m n n 1 n 2 m n 1 m + m + m + ........ + m = m 1 Hence or otherwise prove that, [JEE 2000 (Mains), 6M] n n 1 n 2 m n 2 m + 2 m + 3 m + ........ + (n m + 1) m = m 2 . GF JIm 10 20 p H K3 . 0) [JEE 02(Screening ), 3M] The sum i0 i m i (where if p < q is maximum when m is - q (A) 5 (B) 10 (C) 15 (D) 20 4 . ( a ) Coefficient of t24 in the expansion of (1 + t2)12 (1 + t12) (1 + t24) is - [JEE 03, Screening , 3M out of 60] (A) 12C + 2 (B) 12C + 1 (C) 12C (D) none 6 6 6 [JEE 03, Mains 2M out of 60] ( b ) If n and k are positive integers, show that 2k n n 2 k 1 n n 1 2 k 2 n n 2 .....( 1 ) k n nk n 0 k 1 k 1 2 k 2 k 0 k 5. If n, r N and n–1C = (k2 – 3) (nC ), then k lies in the interval - r r+1 [JEE 04, Screening, 3M out of 84] (A) 3, 3 (B) (2, ) (C) 3, (D) 3, 2 6. The value of FHG300JKI FHG1300IKJ GFH310KJI GHF3101IKJ FGH320JKI GHF3102IKJ HGF IJK30 HGF3300IKJ , is where GFHrnIKJ nCr NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#11\\Eng\\01.BINOMIAL.p65 .......... 20 [JEE 05, Screening, 3M out of 84] (A) 30C10 (B) 60C20 (C) 31C11 or 31C10 (D) 30C11 7 . For r = 0, 1,....,10, let Ar, Br and Cr denote, respectively, the coefficient of xr in the expansions of (1+x)10, 10 [JEE 10, 5M, –2M] (1 + x)20 and (1 + x)30. Then A r (B10Br C10 A r ) is equal to - r 1 (B) 2 (A) B10 – C10 A 10 B 10 C10 A10 (C) 0 (D) C10 – B10 8 . The coefficients of three consecutive terms of (1 + x)n+5 are in the ratio 5 : 10 : 14. Then n = [JEE-Advanced 2013, 4, (–1)] PREVIOUS YEARS QUESTIONS ANSWER KEY EXERCISE-5 [B] 1. (a) D ; (b) B 3. C 4. (a) A 5. D 6. A 7. D 8. 6 26 E
JEE-Mathematics CIRCLE 1. (A) DEFINITION : A circle is the locus of a point which moves in a plane in such a way that its distance from a fixed point (in the same given plane) remains constant. The fixed point is called the centre of the circle and the constant distance is called the radius of the circle. Equation of a circle : The curve traced by the moving point is called its circumference i.e. the equation of any circle is satisfied by co-ordinates of all points on its circumference. or The equation of the circle means the equation of its circumference. or It is the set of all points lying on the circumference of the circle. Chord and diameter - the line joining any two points on the circumference is called a P CQ chord. If any chord passing through its centre is called its diameter. AB AB = chord, PQ = diameter C = centre (B) BASIC THEOREMS & RESULTS OF CIRCLES : ( a ) Concentric circles : Circles having same centre. ( b ) Congruent circles : Iff their radii are equal. ( c ) Congruent arcs : Iff they have same degree measure at the centre. Theorem 1 : (i) If two arcs of a circle (or of congruent circles) are congruent, the corresponding chords are equal. Converse : If two chords of a circle are equal then their corresponding arcs are congruent. (ii) Equal chords of a circle (or of congruent circles) subtend equal angles at the centre. Converse : If the angle subtended by two chords of a circle (or of congruent circles) at the centre are equal, the chords are equal. Theorem 2 : (i) The perpendicular from the centre of a circle to a chord bisects the chord. Converse : The line joining the mid point of a chord to the centre of a circle is perpendicular to the chord. (ii) Perpendicular bisectors of two chords of a circle intersect at its centre. Theorem 3 : (i) There is one and only one circle passing through three non collinear points. (ii) If two circles intersects in two points, then the line joining the centres is perpendicular bisector of Node6\\E_NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#05\\Eng\\02 CIRCLE.p65 common chords. Theorem 4 : (i) Equal chords of a circle (or of congruent circles) are equidistant from the centre. Converse : Chords of a circle (or of congruent circles) which are equidistant from the centre are equal in length. (ii) If two equal chords are drawn from a point on the circle, then the centre of circle will lie on angle bisector of these two chords. (iii) Of any two chords of a circle larger will be near to centre. Theorem 5 : O (i) The degree measure of an arc or angle subtended by an arc at the centre is 2 double the angle subtended by it at any point of alternate segment. (ii) Angle in the same segment of a circle are equal. 44 E
JEE-Mathematics (iii) The angle in a semi circle is right angle. Converse : The arc of a circle subtending a right angle in alternate segment is semi circle. Theorem 6 : Any angle subtended by a minor arc in the alternate segment is acute and any angle subtended by a major arc in the alternate segment is obtuse. Theorem 7 : If a line segment joining two points subtends equal angles at two other points lying on the same side of the line segment, the four points are concyclic, i.e. lie on the same circle. (d) Cyclic Quadrilaterals : A quadrilateral is called a cyclic quadrilateral if its all vertices lie on a circle. Theorem 1 : The sum of either pair of opposite angles of a cyclic quadrilateral is 180° OR The opposite angles of a cyclic quadrilateral are supplementary. Converse : If the sum of any pair of opposite angle of a quadrilateral is 180°, then the quadrilateral is cyclic. Theorem 2 : If a side of a cyclic quadrilateral is produced, then the exterior angle is equal to the interior opposite angle. Theorem 3 : C The internal angle bisectors of a cyclic quadrilateral form a quadrilateral which is also cyclic. D Theorem 4 : If two sides of a cyclic quadrilateral are parallel then the remaining two sides are P QS R AB equal and the diagonals are also equal. OR A cyclic trapezium is isosceles and its diagonals are equal. Converse : If two non-parallel sides of a trapezium are equal, then it is cyclic. OR An isosceles trapezium is always cyclic. Theorem 5 : When the opposite sides of cyclic quadrilateral (provided that they are not parallel) are produced, then Node6\\E_NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#05\\Eng\\02 CIRCLE.p65 the exterior angle bisectors intersect at right angle. (C) TANGENTS TO A CIRCLE : Theorem 1 : A tangent to a circle is perpendicular to the radius through the point of contact. Converse : A line drawn through the end point of a radius and perpendicular to it is a tangent to the circle. Theorem 2 : If two tangents are drawn to a circle from an external point, then : (i) they are equal. (ii) they subtend equal angles at the centre, (iii) they are equally inclined to the segment, joining the centre to that point. Theorem 3 : If two chords of a circle intersect inside or outside the circle A DA B O P when produced, the rectangle formed by the two segments D of one chord is equal in area to the rectangle formed by the P CB two segments of the other chord. C PA × PB = PC × PD 45 E
JEE-Mathematics Theorem 4 : B If PAB is a secant to a circle intersecting the circle at A and B and PT is A tangent segment, then PA × PB = PT2 PO OR Area of the rectangle formed by the two segments of a chord is equal to T the area of the square of side equal to the length of the tangent from the point on the circle. E Theorem 5 : C If a chord is drawn through the point of contact of a tangent to a circle, B then the angles which this chord makes with the given tangent are equal O respectively to the angles formed in the corresponding alternate segments. D BAQ = ACB and BAP = ADB P AQ Converse : If a line is drawn through an end point of a chord of a circle so that the angle formed with the chord is equal to the angle subtended by the chord in the alternate segment, then the line is a tangent to the circle. 2 . STANDARD EQUATIONS OF THE CIRCLE : (a) Central Form : If (h, k) is the centre and r is the radius of the circle then its equation is (x–h)2 + (y–k)2 = r2 Special Cases : (i) If centre is origin (0,0) and radius is 'r' then equation of circle is x2 + y2 = r2 and this is called the standard form. (ii) If radius of circle is zero then equation of circle is (x – h)2 + (y – k)2 = 0. Such circle is called zero circle or point circle. y (iii) When circle touches x-axis then equation of the circle is (x–h)2 + (y–k)2 = k2. (h,k) C k x 0 Touching x-axis (iv) When circle touches y-axis then equation of circle is y (x–h)2 + (y–k)2 = h2 . C(h,k) h (v) When circle touches both the axes (x-axis and y-axis) then equation of x Node6\\E_NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#05\\Eng\\02 CIRCLE.p65 0 Touching y-axis y circle (x–h)2 + (y–h)2 = h2. C(h,h) hh x 0 Touching x-axis and y-axis (vi) When circle passes through the origin and centre of the circle is (h,k) y then radius h2 k2 r and intercept cut on x-axis OP =2h, Q x and intercept cut on y-axis is OQ = 2k and equation of circle is (0,2k) C(h,k) (x–h)2 + (y–k)2 = h2 + k2 or x2 + y2 – 2hx – 2ky = 0 k (2h,0) OP Note : Centre of the circle may exist in any quadrant hence for general cases use ± sign before h & k. 46 E
JEE-Mathematics (b) General equation of circle x2 + y2 + 2gx + 2fy + c = 0. where g,f,c are constants and centre is (–g,–f) i.e. co efficien t of x coefficient of y and radius r g2 f2 c 2 , 2 Note : (i) If (g2 + f2 – c) > 0, then r is real and positive and the circle is a real circle. (ii) If (g2 + f2 – c) = 0, then radius r = 0 and circle is a point circle. (iii) If (g2 + f2 –c)<0, then r is imaginary then circle is also an imaginary circle with real centre. (iv) x2 + y2 + 2gx + 2fy + c = 0, has three constants and to get the equation of the circle at least three conditions should be known A unique circle passes through three non collinear points. (v) The general second degree in x and y, ax2 + by2 + 2hxy + 2gx + 2fy + c = 0 represents a circle if : • coefficient of x2 = coefficient of y2 or a = b 0 • coefficient of xy = 0 or h = 0 • (g2 + f2 – c) 0 (for a real circle) (c) Intercepts cut by the circle on axes : The intercepts cut by the circle x2 + y2 + 2gx + 2fy + c =0 on : (i) x-axis = 2 g2 – c (ii) y-axis = 2 f2 – c Note : (i) If the circle cuts the x-axis at two distinct point, then g2 – c > 0 (ii) If the cirlce cuts the y-axis at two distinct point, then f2 – c > 0 (iii) If circle touches x-axis then g2 = c. (iv) If circle touches y-axis then f2 = c. (v) Circle lies completely above or below the x-axis then g2 < c. (vi) Circle lies completely to the right or left to the y-axis, then f2 < c. (vii) Intercept cut by a line on the circle x2 + y2 + 2gx + 2fy+c=0 or length of O aP chord of the circle 2 a2 P2 where a is the radius and P is the length of AC B perpendicular from the centre to the chord. (d) Equation of circle in diameter form : P(x,y) Node6\\E_NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#05\\Eng\\02 CIRCLE.p65 If A(x ,y ) and B(x ,y ) are the end points of the diameter of the circle (x1,y1) A C B(x2,y2) 11 22 and P(x,y) is the point other then A and B on the circle then from geometry we know that APB = 90°. (Slope of PA) × (Slope of PB) = –1 y y1 y y2 = – 1 x x1 x x2 (x–x ) (x–x )+(y–y )(y–y ) = 0 1 2 12 Note : This will be the circle of least radius passing through (x , y ) and (x , y ) 11 22 (e) Equation of circle in parametric forms : (i) The parametric equation of the circle x2+y2 = r2 are x = r cos, y = r sin ; [0, 2) and (r cos , r sin ) are called the parametric co-ordinates. (ii) The parametric equation of the circle (x – h)2 + (y – k)2 = r2 is x = h + r cos, y = k + r sin where is parameter. (iii) The parametric equation of the circle x2 + y2 + 2gx + 2fy + c = 0 are x = – g + g 2 + f 2 – c cos, y = –f + g 2 + f 2 – c sin where is parameter. E 47
JEE-Mathematics Note : Equation of a straight line joining two point on the circle x2 + y2 = a2 is x cos + y sin = a cos . 222 Illustration 1 : Find the centre and the radius of the circles Solution : (a) 3x2 + 3y2 – 8x – 10y + 3 = 0 (b) x2 + y2 + 2x sin + 2y cos – 8 = 0 (c) 2x2 + xy + 2y2 + ( – 4)x + 6y – 5 = 0, for some . (a) We rewrite the given equation as x2 + y2 – 8 x 10 y 1 0 g = – 4 , f = – 5 , c = 1 33 33 Hence the centre is 4 , 5 and the radius is 16 25 1 32 4 2 units 3 3 99 93 (b) x2 + y2 + 2x sin + 2ycos – 8 = 0. Centre of this circle is (–sin, – cos) Radius = sin2 cos2 8 1 8 3 units (c) 2x2 + xy + 2y2 + ( – 4)x + 6y – 5 = 0 We rewrite the equation as x2 xy y2 4 x 3y 5 0 ........ (i) 2 2 2 Since, there is no term of xy in the equation of circle 5 = 0 = 0 So, equation (i) reduces to x2 + y2 – 2x + 3y 0 2 2 centre is 1, 3 Radius = 95 23 2 1 units. 42 2 Illustration 2 : If the lines 3x – 4y + 4 = 0 and 6x – 8y – 7 = 0 are tangents to a circle, then the radius of the circle Solution : is - (A) 3/2 (B) 3/4 (C) 1/10 (D) 1/20 The diameter of the circle is perpendicular distance between the parallel lines (tangents) 3x – 4y + 4 = 0 and 3x – 4y – 7 = 0 and so it is equal to 4 7/2 3 . 2 9 16 2 3 Ans. (B) Node6\\E_NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#05\\Eng\\02 CIRCLE.p65 Hence radius is . Illustration 3 : 4 If y = 2x + m is a diameter to the circle x2 + y2 + 3x + 4y – 1 = 0, then find m Solution : Centre of circle = (–3/2 , –2). This lies on diameter y = 2x + m – 2 = (–3/2) × 2 + m m = 1 Illustration 4 : The equation of a circle which passes through the point ( 1 , –2) and ( 4 , –3) and whose centre lies on the line 3x + 4y = 7 is (A) 15 ( x2 + y2) – 94x + 18y – 55 = 0 (B) 15 ( x2 + y2) – 94x + 18y + 55 = 0 (C) 15 ( x2 + y2) + 94x – 18y + 55 = 0 (D) none of these Solution : Let the circle be x2 + y2 + 2gx + 2fy + c = 0 ..... (i) Hence, substituting the points, ( 1, –2) and ( 4 , –3) in equation (i) 5 + 2g – 4f + c = 0 ..... (ii) 25 + 8g – 6f + c = 0 ..... (iii) centre ( –g , –f) lies on line 3x + 4y = 7 E 48
JEE-Mathematics Hence –3g –4f = 7 solving for g, f,c, we get 47 , f 9 , c 55 Here g = 15 15 15 Hence the equation is 15 ( x2 + y2 ) –94x + 18y + 55 = 0 Ans. (B) Illustration 5 : A circle has radius equal to 3 units and its centre lies on the line y = x – 1. Find the equation of the Solution : circle if it passes through (7, 3). Let the centre of the circle be ( ). It lies on the line y = x – 1 – 1. Hence the centre is ( –1). The equation of the circle is (x – )2 + (y – + 1)2 = 9 It passes through (7, 3) (7 – )2 + (4 – )2 = 9 22 – 22 + 56 = 0 2 – 11 + 28 = 0 ( – 4)( – 7) = 0 = 4, 7 Hence the required equations are Ans. x2 + y2 – 8x – 6y + 16 = 0 and x2 + y2 – 14x – 12y + 76 = 0. Do yourself - 1 : ( i ) Find the centre and radius of the circle 2x2 + 2y2 = 3x – 5y + 7 ( i i ) Find the equation of the circle whose centre is the point of intersection of the lines 2x – 3y + 4 = 0 & 3x + 4y – 5 = 0 and passes through the origin. (i i i ) Find the parametric form of the equation of the circle x2 + y2 + px + py = 0 ( i v ) Find the equation of the circle the end points of whose diameter are the centres of the circles x2 + y2 + 16x – 14y = 1 & x2 + y2 – 4x + 10y = 2 3 . POSITION OF A POINT W.R.T CIRCLE : ( a ) Let the circle is x2 + y2 + 2gx + 2fy + c = 0 and the point is (x ,y ) then - 11 Point (x ,y ) lies out side the circle or on the circle or inside the circle according as 11 x 2 + y 2 + 2gx +2fy + c >, =, < 0 or S >, =, < 0 1 1 1 1 1 Node6\\E_NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#05\\Eng\\02 CIRCLE.p65 ( b ) The greatest & the least distance of a point A from a circle with centre C & radius r is AC + r & |AC – r| respectively. 4. POWER OF A POINT W.R.T. CIRCLE : E Theorem : The power of point P(x1, y1) w.r.t. the circle x2 + y2 + 2gx + 2ƒy + c = 0 is S1 where S = x12 y 2 2gx1 2ƒy1 c 1 1 Note : If P outside, inside or on the circle then power of point is positive, negative or zero respectively. T B B1 If from a point P(x , y ), inside or outside the circle, a secant be drawn A 11 P A1 intersecting the circle in two points A & B, then PA . PB = constant. The product PA . PB is called power of point P(x1, y1) w.r.t. the circle S x2 + y2 + 2gx + 2ƒy + c = 0, i.e. for number of secants PA.PB = PA . PB 1 1 = PA . PB = ...... = PT2 = S 1 22 49
JEE-Mathematics Illustration 6 : If P(2, 8) is an interior point of a circle x2 + y2 – 2x + 4y – p = 0 which neither touches nor Solution : intersects the axes, then set for p is - (A) p < –1 (B) p < – 4 (C) p > 96 (D) For internal point p(2, 8), 4 + 64 – 4 + 32 – p < 0 p > 96 and x intercept = 2 1 p therefore 1 + p < 0 p < –1 and y intercept = 2 4 p p < –4 Ans. (D) Do yourself - 2 : ( i ) Find the position of the points (1, 2) & (6, 0) w.r.t. the circle x2 + y2 – 4x + 2y – 11 = 0 ( i i ) Find the greatest and least distance of a point P(7, 3) from circle x2 + y2 – 8x – 6y + 16 = 0. Also find the power of point P w.r.t. circle. 5 . TANGENT LINE OF CIRCLE : When a straight line meet a circle on two coincident points then it is called the tangent of the circle. ( a ) Condition of Tangency : (P>r) (P=r) The line L = 0 touches the circle S = 0 if P the length of the (P<r) rP Tangent perpendicular from the centre to that line and radius of the Secant circle r are equal i.e. P = r. (P=0) Diameter Illustration 7 : Find the range of parameter 'a' for which the variable line y = 2x + a lies between the circles Solution : x2 + y2 – 2x – 2y + 1 = 0 and x2 + y2 – 16x – 2y + 61 = 0 without intersecting or touching either circle. The given circles are C1 : (x – 1)2 + (y – 1)2 = 1 and C2 : (x – 8)2 + (y – 1)2 = 4 The line y – 2x – a = 0 will lie between these circle if centre of the circles lie on opposite sides of the line, i.e. (1 – 2 – a)(1 – 16 – a) < 0 a (–15, –1) | 1 2 a | 1, | 1 16 a| 2 Line wouldn't touch or intersect the circles if, 55 |1 + a| > 5 , |15 + a| > 2 5 Node6\\E_NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#05\\Eng\\02 CIRCLE.p65 a > 5 – 1 or a < – 5 – 1, a > 2 5 – 15 or a < –2 5 – 15 Illustration 8 : Hence common values of 'a' are (2 5 – 15, – 5 –1). The equation of a circle whose centre is (3, –1) and which cuts off a chord of length 6 on the line 2x– 5y+ 18 = 0 (A) (x – 3)2 + (y + 1)2 = 38 (B) (x + 3)2 + (y – 1)2 = 38 (C) (x – 3)2 + (y + 1)2 = 38 (D) none of these Solution : Let AB(= 6) be the chord intercepted by the line 2x – 5y + 18 = 0 from the circle and let CD be the perpendicular drawn from centre (3, –1) to the chord AB. C(3,-1) D 2.3 5(1) 18 29 A i.e., AD = 3, CD = 22 52 B Therefore, CA2 = 32 + ( 29 )2 = 38 Ans. (A) Hence required equation is (x – 3)2 + (y + 1)2 = 38 E 50
JEE-Mathematics Illustration 9 : The area of the triangle formed by line joining the origin to the points of intersection(s) of the line x 5 2 y 3 5 and circle x2 + y2 = 10 is (A) 3 (B) 4 (C) 5 (D) 6 Solution : Length of perpendicular from origin to the line x 5 2 y 3 5 is Q OL 3 5 3 5 5 L 9 O ( 5 )2 22 10 P Radius of the given circle = 10 = OQ = OP 5 x + 2y = 3 5 PQ = 2QL = 2 OQ2 OL2 2 10 5 2 5 Ans. (C) 11 Thus area of OPQ = 2 PQ OL 2 2 5 5 5 (b) Equation of the tangent (T = 0) : (i) Tangent at the point (x ,y ) on the circle x2+ y2 = a2 is xx + yy = a2. 11 11 (ii) (1) The tangent at the point (acos t, asin t) on the circle x2 + y2 = a2 is xcos t + ysin t = a a cos , a sin 2 cos 2 (2) The point of intersection of the tangents at the points P() and Q() is . cos 2 2 (iii) The equation of tangent at the point (x ,y ) on the circle x2 + y2 + 2gx + 2fy + c = 0 is 11 xx + yy + g(x + x ) + f(y + y ) + c = 0 11 1 1 (iv) If line y = mx + c is a straight line touching the circle x2 + y2 = a2, then c = ± a 1 m2 and contact points are am a or a 2m , a2 and equation of tangent is , + m2 c c 1 1+ m2 y = mx ± a 1 + m 2 (v) The equation of tangent with slope m of the circle (x – h)2 + (y – k)2 = a2 is (y – k) = m(x – h) ± a 1 + m 2 Note : To get the equation of tangent at the point (x y ) on any second degree curve we replace xx in 11 1 place of x2, yy1 in place of y2, x x1 in place of x, y y1 in place of y, xy1 yx1 in place of xy 2 2 2 Node6\\E_NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#05\\Eng\\02 CIRCLE.p65 and c in place of c. (c) Length of tangent ( S1 ) : T The length of tangent drawn from point (x ,y ) out side the circle P(x1,y1) 11 S x2 + y2 + 2gx + 2fy + c = 0 is, PT= S1 = x12 y12 2gx1 2fy1 c Note : When we use this formula the coefficient of x2 and y2 must be 1. (d) Equation of Pair of tangents (SS = T2) : 1 Let the equation of circle S x2 + y2 = a2 and P(x ,y ) is any point Q 11 M outside the circle. From the point we can draw two real and distinct (0,0) tangent PQ & PR and combine equation of pair of tangents is - P R (x1,y1) (x2 + y2 – a2) (x 2 + y 2 – a2) = (xx + yy – a2)2 or 11 11 SS = T2 51 1 E
JEE-Mathematics Illustration 10 : Let A be the centre of the circle x2 + y2 – 2x – 4y – 20 = 0 and B(1, 7) and D(4, –2) are points on the circle then, if tangents be drawn at B and D, which meet at C, then area of quadrilateral ABCD is - (A) 150 (B) 75 (C) 75/2 (D) none of these Solution : B (1, 7) (1, 2) C A (16, 7) D (4, –2) Here centre A(1, 2) and Tangent at (1, 7) is x.1 + y.7 – 1(x + 1) – 2(y + 7) – 20 = 0 or y = 7 .......... (i) Tangent at D(4, –2) is 3x – 4y – 20 = 0 .......... (ii) Solving (i) and (ii), C is (16, 7) Area ABCD = AB × BC = 5 × 15 = 75 units. Ans. (B) Do yourself - 3 : ( i ) Find the equation of tangent to the circle x2 + y2 – 2ax = 0 at the point (a(1 + cos), asin). ( i i ) Find the equations of tangents to the circle x2 + y2 – 6x + 4y – 12 = 0 which are parallel to the line 4x – 3y + 6 = 0 (iii) Find the equation of the tangents to the circle x2 + y2 = 4 which are perpendicular to the line 12x – 5y + 9 = 0. Also find the points of contact. ( i v ) Find the value of 'c' if the line y = c is a tangent to the circle x2 + y2 – 2x + 2y – 2 = 0 at the point (1, 1) 6 . NORMAL OF CIRCLE : Normal at a point is the straight line which is perpendicular to the tangent at the point of contact. Note : Normal at point of the circle passes through the centre of the circle. ( a ) Equation of normal at point (x ,y ) of circle x2 + y2 + 2gx + 2fy + c = 0 is N (–g, –f) 11 y– y = y1 + f (x - x1 ) PT Node6\\E_NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#05\\Eng\\02 CIRCLE.p65 1 x1 + g (x1,y1) (b) The equation of normal on any point (x ,y ) of circle x2 + y2 = a2 is y y1 - 11 x x1 ( c ) If x2 + y2 = a2 is the equation of the circle then at any point 't' of this circle (a cos t, a sint), the equation of normal is xsint – ycost = 0. Illustration 11 : Find the equation of the normal to the circle x2 + y2 – 5x + 2y – 48 = 0 at the point (5, 6). Solution : Since normal to the circle always passes through the centre so equation of the normal will be the line passing through (5, 6) & 5 , 1 2 i.e. y + 1 = 5 7 2 x 5 5y 5 14x 35 / 2 14x – 5y – 40 = 0 Ans. 52 E
JEE-Mathematics Illustration 12 : If the straight line ax + by = 2; a, b 0 touches the circle x2 + y2 – 2x = 3 and is normal to the circle x2 + y2 – 4y = 6, then the values of a and b are respectively (A) 1, –1 (B) 1, 2 4 (D) 2, 1 (C) , 1 3 Solution : Given x2 + y2 – 2x = 3 centre is (1, 0) and radius is 2 Given x2 + y2 – 4y = 6 centre is (0, 2) and radius is 10 . Since line ax + by = 2 touches the first circle | a(1) b(0) 2| 2 or |(a – 2)| = [2 a2 b2 ] ......... (i) a2 b2 Also the given line is normal to the second circle. Hence it will pass through the centre of the second circle. a(0) + b(2) = 2 or 2b = 2 or b = 1 Putting this value in equation (i) we get |a – 2| = 2 a2 12 or (a – 2)2 = 4(a2 + 1) 4 or a2 + 4 – 4a = 4a2 + 4 or 3a2 + 4a = 0 or a (3a + 4) = 0 or a = 0, (a 0) 3 values of a and b are 4 , 1 . Ans. (C) 3 Illustration 13 : Find the equation of a circle having the lines x2 + 2xy + 3x + 6y = 0 as its normal and having size just sufficient to contain the circle x(x – 4) + y(y – 3) = 0. Solution : Pair of normals are (x + 2y)(x + 3) = 0 Normals are x + 2y = 0, x + 3 = 0. Point of intersection of normals is the centre of required circle i.e. C1(–3, 3/2) and centre of given 95 circle is C2(2, 3/2) and radius r2 = 4 42 Let r1 be the radius of required circle r1 = C1C2 + r2 = (3 2)2 3 32 5 15 2 2 2 2 Hence equation of required circle is x2 + y2 + 6x – 3y – 45 = 0 Node6\\E_NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#05\\Eng\\02 CIRCLE.p65 Do yourself - 4 : ( i ) Find the equation of the normal to the circle x2 + y2 = 2x, which is parallel to the line x + 2y = 3. 7. CHORD OF CONTACT (T = 0) : E A line joining the two points of contacts of two tangents drawn from a point out T1 R side the circle, is called chord of contact of that point. C L If two tangents PT & PT are drawn from the point P (x , y ) to the circle P(x1,y1) 12 11 S x2 + y2 + 2gx + 2fy + c = 0 , then the equation of the chord of contact T T is : T2 12 xx + yy + g (x + x ) + f (y + y ) + c = 0 (i.e. T = 0 same as equation of tangent). 1 1 1 1 Remember : 2LR (a) Length of chord of contact T T = R 2 L2 . 1 2 R L3 ( b ) Area of the triangle formed by the pair of the tangents & its chord of contact = R 2 L2 , where R is the radius of the circle & L is the length of the tangent from (x1, y1) on S = 0. 53
JEE-Mathematics (c) Angle between the pair of tangents from P(x , y ) = tan 1 2RL 11 L2 R2 ( d ) Equation of the circle circumscribing the triangle PT T or quadrilateral CT PT is : 1 2 12 (x - x ) (x + g) + (y – y ) (y + f) = 0. 11 ( e ) The joint equation of a pair of tangents drawn from the point A (x , y ) to the circle 1 1 x2 + y2 + 2gx + 2fy + c = 0 is : SS = T². 1 Where S x2 + y2 + 2gx + 2fy + c ; S x ² + y ² + 2gx + 2fy + c 1 1 1 1 1 T xx + yy + g(x + x ) + f(y + y ) + c. 1 1 1 1 Illustration 14 : The chord of contact of tangents drawn from a point on the circle x2 + y2 = a2 to the circle x2 + y2 = b2 touches the circle x2 + y2 = c2. Show that a, b, c are in GP. Solution : Let P(acos, asin) be a point on the circle x2 + y2 = a2. T Then equation of chord of contact of tangents drawn from P P(acos, asin) to the circle x2 + y2 = b2 is axcos + aysin = b2 ..... (i) This touches the circle x2 + y2 = c2 ..... (ii) R Length of perpendicular from (0, 0) to (i) = radius of (ii) x2+y2=c2 x2+y2=b2 | 0 0 b2 | x2+y2=a2 c (a2 cos2 a2 sin2 ) or b2 = ac a, b, c are in GP. Do yourself - 5 : (i) Find the equation of the chord of contact of the point (1, 2) with respect to the circle x2 + y2 + 2x + 3y + 1 = 0 ( i i ) Tangents are drawn from the point P(4, 6) to the circle x2 + y2 = 25. Find the area of the triangle formed by them and their chord of contact. 8 . EQUATION OF THE CHORD WITH A GIVEN MIDDLE POINT (T = S ) : 1 The equation of the chord of the circle S x2 + y2 + 2gx + 2fy + c = 0 in terms of its mid point M (x , y) 1 1 is y y = x1 g (x x ). This on simplification can be put in the form 1 y1 f 1 xx + yy + g (x + x ) + f (y + y ) + c = x 2 + y 2 + 2gx + 2fy + c which is designated by T = S . 1 1 1 1 11 1 1 1 Note that : The shortest chord of a circle passing through a point ‘M’ inside the circle, is one chord whose middle point is M. Node6\\E_NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#05\\Eng\\02 CIRCLE.p65 Illustration 15 : Find the locus of middle points of chords of the circle x2 + y2 = a2, which subtend right angle at the point (c, 0). Solution : Let N(h, k) be the middle point of any chord AB, y which subtend a right angle at P(c, 0). A Since APB = 90° (h,k) x NA = NB = NP N (since distance of the vertices from middle point of the hypotenuse are equal) x' or (NA)2 = (NB)2 = (h – c)2 + (k – 0)2 ..... (i) P(c, 0) O B But also BNO = 90° y' (OB)2 = (ON)2 + (NB)2 –(NB)2 = (ON)2 – (OB)2 –[(h – c)2 + (k – 0)2] = (h2 + k2) – a2 or 2(h2 + k2) – 2ch + c2 – a2 = 0 Locus of N(h, k) is 2(x2 + y2) – 2cx + c2 – a2 = 0 Ans. 54 E
JEE-Mathematics Illustration 16 : Let a circle be given by 2x(x – a) + y(2y – b) = 0 (a 0, b 0) Find the condition on a and b if two chords, each bisected by the x-axis, can be drawn to the circle from (a, b/2). Solution : The given circle is 2x(x – a) + y(2y – b) = 0 or x2 + y2 – ax – by/2 = 0 Let AB be the chord which is bisected by x-axis at a point M. Let its co-ordinates be M(h, 0). and S x2 + y2 – ax – by/2 = 0 Equation of chord AB is T = S1 hx + 0 – a (x h) b (y 0) h2 0 ah 0 24 a b2 = h2 – ah h2 – 3ah a2 b2 0 Since its passes through (a, b/2) we have ah – (a + h) – 28 2 28 Now there are two chords bisected by the x-axis, so there must be two distinct real roots of h. B2 – 4AC > 0 3a 2 a2 b2 0 a2 > 2b2. Ans. 2 4.1. 2 8 Do yourself - 6 : ( i ) Find the equation of the chord of x2 + y2 – 6x + 10 – a = 0 which is bisected at (–2, 4). ( i i ) Find the locus of mid point of chord of x2 + y2 + 2gx + 2ƒy + c = 0 that pass through the origin. 9. DIRECTOR CIRCLE : The locus of point of intersection of two perpendicular tangents to a circle is called director circle. Let P(h,k) is the point of intersection of two tangents drawn on the circle x2 + y2 = a2. Then the equation of the pair of tangents is SS = T2 1 i.e. (x2 + y2 – a2) (h2 + k2 – a2) = (hx + ky – a2)2 As lines are perpendicular to each other then, coefficient of x2 + coefficient of y2 = 0 [(h2 +k2 – a2)–h2] + [(h2 + k2 – a2)– k2] = 0 h2 + k2 = 2a2 locus of (h,k) is x2 + y2 = 2a2 which is the equation of the director circle. director circle is a concentric circle whose radius is 2 times the radius of the circle. Note : The director circle of x2 + y2 + 2gx + 2fy + c = 0 is x2 + y2 + 2gx + 2fy + 2c– g2 – f2 = 0 Node6\\E_NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#05\\Eng\\02 CIRCLE.p65 Illustration 17 : Let P be any moving point on the circle x2 + y2 – 2x = 1, from this point chord of contact is drawn w.r.t. the circle x2 + y2 – 2x = 0. Find the locus of the circumcentre of the triangle CAB, C being centre of the circle and A, B are the points of contact. Solution : The two circles are (x – 1)2 + y2 = 1 (x – 1)2 + y2 = 2 ......... (i) ......... (ii) So the second circle is the director circle of the first. So APB = /2 Also ACB = /2 Now circumcentre of the right angled triangle CAB would lie on the mid point of AB So let the point be M (h, k) 1 P Now, CM = CBsin45° = A B 2 M So, ( h – 1)2 + k 2 = 1 2 C 2 So, locus of M is (x – 1)2 + y2 = 1 . 2 E 55
JEE-Mathematics Do yourself - 7 : ( i ) Find the equation of the director circle of the circle (x – h)2 + (y – k)2 = a2. ( i i ) If the angle between the tangents drawn to x2 + y2 + 4x + 8y + c = 0 from (0, 0) is , then find value 2 of 'c' (i i i ) If two tangents are drawn from a point on the circle x2 + y2 = 50 to the circle x2 + y2 = 25, then find the angle between the tangents. 10. POLE AND POLAR : R (h,k) Let any straight line through the given point A(x ,y ) intersect the given circle S =0 in A (x1,Py1 ) Q 11 two points P and Q and if the tangent of the circle at P and Q meet at the point R then locus of point R is called polar of the point A and point A is called the pole, with respect to the given circle. ( a ) The equation of the polar of point (x ,y ) w.r.t. circle x2 + y2 = a2 (T = 0). S (h,k) 11 (x1,y1) R Let PQR is a chord which passes through the point P(x ,y ) which intersects the PQ 11 circle at points Q and R and the tangents are drawn at points Q and R meet at point S(h,k) then equation of QR the chord of contact is x h + y k= a2 11 locus of point S(h,k) is xx + yy = a2 which is the equation of the polar. 11 Note : (i) The equation of the polar is the T=0, so the polar of point (x ,y ) w.r.t circle 11 x2 + y2 + 2gx + 2fy + c = 0 is xx + yy + g(x + x ) + f(y + y )+c = 0 11 1 1 (ii) If point is outside the circle then equation of polar and chord of contact is same. So the chord of contact is polar. (iii) If point is inside the circle then chord of contact does not exist but polar exists. (iv) If point lies on the circle then polar , chord of contact and tangent on that point are same. (v) If the polar of P w.r.t. a circle passes through the point Q, then the polar of point Q will pass through P and hence P & Q are conjugate points of each other w.r.t. the given circle. (vi) If pole of a line w.r.t. a circle lies on second line. Then pole of second line lies on first line and hence both lines are conjugate lines of each other w.r.t. the given circle. (vii) If O be the centre of a circle and P be any point, then OP is perpendicular to the polar of P. (viii) If O be the centre of a circle and P any point, then if OP (produce, if necessary) meet the polar of P in Q, then OP. OQ = (radius)2 Node6\\E_NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#05\\Eng\\02 CIRCLE.p65 (b) Pole of a given line with respect to a circle To find the pole of a line we assume the coordinates of the pole then from these coordinates we find the polar. This polar and given line represent the same line. Then by comparing the coefficients of similar terms we can get the coordinates of the pole. The pole of x + my + n = 0 a2 ma2 w.r.t. circle x2 + y2 = a2 will be n , n 1 1 . FA MILY OF CIRCLES : ( a ) The equation of the family of circles passing through the S1 S2 SL points of intersection of two circles E S = 0 & S = 0 is : S + K S = 0 (K –1). 1 2 1 2 ( b ) The equation of the family of circles passing through the point of intersection of a circle S = 0 & a line L = 0 is given by S + KL = 0. 56
JEE-Mathematics ( c ) The equation of a family of circles passing through two given points (x , y ) & (x , y ) can be written in the form : ACB 1 1 2 2 x y1 (x1,y1) (x2,y2) (x x ) (x x2) + (y y ) (y y ) + K x1 y1 1 = 0 where K is a parameter. 1 1 2 x2 y2 1 (d) The equation of a family of circles touching a fixed line y y = m (x x) 1 1 at the fixed point (x , y ) is (x x )2 + (y y )2 + K [y y m (x x )] = 0, (x1,y1) 1 1 1 1 1 1 where K is a parameter. ( e ) Family of circles circumscribing a triangle whose sides are given by 1 2 L = 0 ; L = 0 & L = 0 is given by ; LL + LL + LL = 0 1 2 3 12 23 31 3 provided coefficient of xy = 0 & coefficient of x2 = coefficient of y2. ( f ) Equation of circle circumscribing a quadrilateral whose sides in order are 2 represented by the lines L = 0, L = 0, L = 0 & L = 0 is LL + LL = 0 1 3 1 2 3 4 13 24 4 provided coefficient of x2 = coefficient of y2 and coefficient of xy = 0. Illustration 18 : The equation of the circle through the points of intersection of x2 + y2 – 1 = 0, x2 + y2 – 2x – 4y + 1 = 0 and touching the line x + 2y = 0, is - (A) x2 + y2 + x + 2y = 0 (B) x2 + y2 – x + 20 = 0 (C) x2 + y2 – x – 2y = 0 (D) 2(x2 + y2) – x – 2y = 0 Solution : Family of circles is x2 + y2 – 2x – 4y + 1 + (x2 + y2 – 1) = 0 (1 + ) x2 + (1 + ) y2 – 2x – 4y + (1 – ) = 0 2 4 1 x2 + y2 x y 0 1 1 1 Centre is 1, 2 and radius = 1 2 2 2 1 4 2 . 1 1 |1 | 1 1 1 Node6\\E_NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#05\\Eng\\02 CIRCLE.p65 Since it touches the line x + 2y = 0, hence Radius = Perpendicular distance from centre to the line. i.e., 1 2 2 4 2 5 4 2 = ± 1 1 1 |1 | 12 22 = –1 cannot be possible in case of circle. So = 1. Ans. (C) Thus, we get the equation of circle. Do yourself - 8 : ( i ) Prove that the polar of a given point with respect to any one of circles x2 + y2 – 2kx + c2 = 0, where k is a variable, always passes through a fixed point, whatever be the value of k. ( i i ) Find the equation of the circle passing through the points of intersection of the circle x2 + y2 – 6x + 2y + 4 = 0 & x2 + y2 + 2x – 4y – 6 = 0 and with its centre on the line y = x. (i i i ) Find the equation of the circle through the points of intersection of the circles x2 + y2 + 2x + 3y – 7 = 0 and x2 + y2 + 3x – 2y – 1 = 0 and passing through the point (1, 2). E 57
JEE-Mathematics 1 2 . DIRECT AND TR ANSVERSE COMMON TANGENTS : Let two circles having centre C and C and radii, r and r and C C is the distance between their centres then : 12 12 12 ( a) Both circles will touch : ( i ) Externally if C C = r + r i.e. the distance between their P T 12 1 2 C1 C2 centres is equal to sum of their radii and point P & T divides C C in the ratio r : r (internally & externally respectively). 12 12 In this case there are three common tangents. ( i i ) Internally if C C = |r –r | i.e. the distance between their centres is equal 12 12 to difference between their radii and point P divides C C in the ratio r : r C1 C2 P 12 1 2 externally and in this case there will be only one common tangent. (b) The circles will intersect : when |r – r | < C C < r + r in this case there are 12 12 1 2 two common tangents. C1 C2 (c) The circles will not intersect : (i) One circle will lie inside the other circle if C C < | r –r | In this case there 12 12 will be no common tangent. (ii) When circle are apart from each other then C C >r +r and in this case there 12 1 2 will be four common tangents. Lines PQ and RS are R called transverse or indirect or internal common A C1 tangents and these lines meet line C C on T and T divides Q B 12 1 1 T1 C2 T2 the line C C in the ratio r : r internally and lines AB & CD D 12 1 2 are called direct or external common tangents. These lines meet C C produced on T . Thus T divides C C C S 12 22 12 P externally in the ratio r : r . 12 Note : Length of direct common tangent = (C1C2 )2 (r1 r2 )2 Length of transverse common tangent = (C1C2 )2 (r1 r2 )2 Illu stration 19 : Prove that the circles x2 + y2 + 2ax + c2 = 0 and x2 + y2 + 2by + c2 = 0 touch each other, Solution : if 1 1 1 . x2 + y2 + 2ax + c2 = 0 ....... (i) Node6\\E_NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#05\\Eng\\02 CIRCLE.p65 a2 b2 c2 x2 + y2 + 2by + c2 = 0 ....... (ii) Given circles are and Let C1 and C2 be the centres of circles (i) and (ii), respectively and r1 and r2 be their radii, then C1 = (–a, 0), C2 = (0, –b), r1 a2 c2 , r2 b2 c2 Here we find the two circles touch each other internally or externally. For touch, |C1C2| = |r1 ± r2| or a2 b2 a2 c2 b2 c2 On squaring a2 + b2 = a2 – c2 + b2 – c2 ± 2 a2 c2 b2 c2 or c2 = ± a2 b2 c2 (a2 b2 ) c 4 Again squaring, c4 = a2b2 – c2(a2 + b2) + c4 or c2(a2 + b2) = a2b2 11 1 or a2 b2 c2 58 E
JEE-Mathematics Do yourself - 9 : ( i ) Two circles with radius 5 touches at the point (1, 2). If the equation of common tangent is 4x + 3y = 10 and one of the circle is x2 + y2 + 6x + 2y – 15 = 0. Find the equation of other circle. ( i i ) Find the number of common tangents to the circles x2 + y2 = 1 and x2 + y2 – 2x – 6y + 6 = 0. 13 . THE ANGLE OF INTERSECTION OF TWO CIRCLES : Definition : The angle between the tangents of two circles at the point of intersection of the two circles is called angle of intersection of two circles. If two circles are S x2 + y2 + 2g x + 2f y + c = 0 1 1 1 1 S x2 + y2 + 2g x + 2f y + c = 0 and is the acute angle between them P 2 2 22 then cos 2g1 g2 2f1 f2 c1 c2 or cos r12 r22 d2 r1 r2 2r1r2 2 g12 f12 c1 g22 f22 c2 C1 d C2 Here r and r are the radii of the circles and d is the distance between their 12 centres. If the angle of intersection of the two circles is a right angle then such circles are called \"Orthogonal circles\" and conditions for the circles to be orthogonal is - 2g g + 2f f = c + c 12 12 1 2 1 4 . RADICAL AXIS OF THE TWO CIRCLES (S – S = 0) : P(h,k) 1 2 ( a ) Definition : The locus of a point, which moves in such a way A B that the length of tangents drawn from it to the circles are equal and is called the radical axis. If two circles are - S x2 + y2 + 2g x + 2f y + c =0 1 1 1 1 S x2 + y2 + 2g x + 2f y + c = 0 Radical axis 2 2 2 2 Let P(h,k) is a point and PA,PB are length of two tangents on the circles from point P, Then from definition - h2 k2 2g1h 2f1k c1 h2 k2 2g2h 2f2k c2 or 2(g –g ) h + 2(f –f ) k + c – c = 0 12 12 12 locus of (h,k) 2x(g –g ) + 2y(f –f )k + c – c = 0 12 12 12 Node6\\E_NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#05\\Eng\\02 CIRCLE.p65 S – S = 0 12 which is the equation of radical axis. Note : (i) To get the equation of the radical axis first of all make the coefficient of x2 and y2 =1 (ii) If circles touch each other then radical axis is the common tangent to both the circles. (iii) When the two circles intersect on real points then common chord is the radical axis of the two circles. (iv) The radical axis of the two circles is perpendicular to the line joining the centre of two circles but not always pass through mid point of it. (v) Radical axis (if exist) bisects common tangent to two circles. (vi) The radical axes of three circles (taking two at a time) meet at a point. (vii) If circles are concentric then the radical axis does not always exist but if circles are not concentric then radical axis always exists. (viii) If two circles are orthogonal to the third circle then radical axis of both circle passes through the centre of the third circle. (ix) A system of circle, every pair of which have the same radical axis, is called a coaxial system of circles. 59 E
JEE-Mathematics (b) Radical centre : The radical centre of three circles is the point from which length of tangents on three circles are equal i.e. the point of intersection of radical axis of the circles is the radical centre of the circles. To get the radical axis of three circles S =0, S =0, S =0 we have to solve any two 1 23 S –S =0, S –S =0, S –S =0 12 23 31 Note : I T1 (i) The circle with centre as radical centre and radius equal to the C1 T2 C2 length of tangent from radical centre to any of the circle, will cut = the three circles orthogonally. = (ii) If three circles are drawn on three sides of a triangle taking them = as diameter then its orthocenter will be its radical centre. III T3 II (iii) Locus of the centre of a variable circle orthogonal to two fixed C3 circles is the radical axis between the two fixed circles. (iv) If two circles are orthogonal, then the polar of a point 'P' on first circle w.r.t. the second circle passes through the point Q which is the other end of the diameter through P. Hence locus of a point which moves such that its polars w.r.t. the circles S = 0 , S = 0 & S = 0 are concurrent 12 3 is a circle which is orthogonal to all the three circles. Illustration 20 : A and B are two fixed points and P moves such that PA = nPB where n 1. Show that locus of P is a circle and for different values of n all the circles have a common radical axis. Solution : Let A (a, 0), B (–a, 0) and P(h, k) so PA = nPB (h – a)2 + k2 = n2[(h + a)2 + k2] (1 – n2)h2 + (1 – n2)k2 – 2ah(1 + n2) + (1 – n2)a2 = 0 h2 + k2 – 2ah 1 n 2 a2 0 1 n 2 Hence locus of P is x2 + y 2 – 2ax 1 n 2 a2 0 , which is a circle of different values of n. 1 n 2 Let n1 and n2 are two different values of n so their radical axis is x = 0 i.e. y-axis. Hence for different Node6\\E_NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#05\\Eng\\02 CIRCLE.p65 values of n the circles have a common radical axis. Illustration 21 : Find the equation of the circle through the points of intersection of the circles x2 + y2 – 4x – 6y – 12 = 0 and x2 + y2 + 6x + 4y – 12 = 0 and cutting the circle x2 + y2 – 2x – 4 = 0 orthogonally. Solution : The equation of the circle through the intersection of the given circles is x2 + y2 – 4x – 6y – 12 + (–10x – 10y) = 0 .......... (i) where (–10x – 10y = 0) is the equation of radical axis for the circle x2 + y2 – 4x – 6y – 12 = 0 and x2 + y2 + 6x + 4y – 12 = 0. Equation (i) can be re-arranged as x2 + y2 – x(10 + 4) – y(10 + 6) – 12 = 0 It cuts the circle x2 + y2 – 2x – 4 = 0 orthogonally. Hence 2gg1 + 2ff1 = c + c1 2(5 + 2)(1) + 2(5 + 3)(0) = – 12 – 4 = – 2 Hence the required circle is E x2 + y2 – 4x – 6y – 12 – 2(–10x – 10y) = 0 i.e., x2 + y2 + 16x + 14y – 12 = 0 60
JEE-Mathematics Illustration 22 : Find the radical centre of circles x2 + y2 + 3x + 2y + 1 = 0, x2 + y2 – x + 6y + 5 = 0 and x2 + y2 + 5x – 8y + 15 = 0. Also find the equation of the circle cutting them orthogonally. Solution : Given circles are S1 x2 + y2 + 3x + 2y + 1 = 0 S2 x2 + y2 – x + 6y + 5 = 0 S3 x2 + y2 + 5x – 8y + 15 = 0 Equations of two radical axes are S1 – S2 4x – 4y – 4 = 0 or x – y – 1 = 0 and S2 – S3 – 6x + 14y – 10 = 0 or 3x – 7y + 5 = 0 Solving them the radical centre is (3, 2). Also, if r is the length of the tangent drawn from the radical centre (3, 2) to any one of the given circles, say S1, we have r = S1 32 22 3.3 2.2 1 27 Hence (3, 2) is the centre and 27 is the radius of the circle intersecting them orthogonally. Its equation is (x – 3)2 + (y – 2)2 = r2 = 27 x2 + y2 – 6x – 4y – 14 = 0 Alternative Method : Let x2 + y2 + 2gx + 2fy + c = 0 be the equation of the circle cutting the given circles orthogonally. 2g 3 +2f(1) = c + 1 or 3g + 2f = c + 1 ........ (i) 2 2g 1 +2f(3) = c + 5 or –g + 6f = c + 5 ........ (ii) 2 5 ........ (iii) and 2g 2 +2f(–4) = c + 15 or 5g – 8f = c + 15 Ans. Solving (i), (ii) and (iii) we get g = –3, f = –2 and c = –14 equation of required circle is x2 + y2 – 6x – 4y – 14 = 0 Node6\\E_NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#05\\Eng\\02 CIRCLE.p65 Do yourself - 10 : ( i ) Find the angle of intersection of two circles S : x2 + y2 – 4x + 6y + 11 = 0 & S' : x2 + y2 – 2x + 8y + 13 = 0 (ii) Find the equation of the radical axis of the circle x2 + y2 – 3x – 4y + 5 = 0 and 3x2 + 3y2 – 7x – 8y +11 = 0 (i i i ) Find the radical centre of three circles described on the three sides 4x – 7y + 10 = 0, x + y – 5 = 0 and 7x + 4y – 15 = 0 of a triangle as diameters. 1 5 . SOME IMPORTANT RESULTS TO REMEMBER : ( a ) If the circle S = 0, bisects the circumference of the circle S = 0, then their common chord will be the 1 2 diameter of the circle S2 = 0. ( b ) The radius of the director circle of a given circle is 2 times the radius of the given circle. ( c ) The locus of the middle point of a chord of a circle subtend a right angle at a given point will be a circle. ( d ) The length of side of an equilateral triangle inscribed in the circle x2 + y2 = a2 is 3 a ( e ) If the lengths of tangents from the points A and B to a circle are 1 and 2 respectively, then if the points A and B are conjugate to each other, then (AB)2 = 21 22 . ( f ) Length of transverse common tangent is less than the length of direct common tangent. E 61
JEE-Mathematics Do yourself - 11 : ( i ) When the circles x2 + y2 + 4x + 6y + 3 = 0 and 2(x2 + y2) + 6x + 4y + c = 0 intersect orthogonally, then find the value of c is ( i i ) Write the condition so that circles x2 + y2 + 2ax + c = 0 and x2 + y2 + 2by + c = 0 touch externally. Miscellaneous Illustrations : Illustration 23 : Find the equation of a circle which passes through the point (2, 0) and whose centre is the limit of the point of intersection of the lines 3x + 5y = 1 and (2 + c)x + 5c2y = 1 as c 1. Solution : Solving the equations (2 + c)x + 5c2y = 1 and 3x + 5y = 1 then (2 + c)x + 5c2 1 3x = 1 or (2 + c)x + c2 (1 – 3x) = 1 5 1 c2 or x (1 c)(1 c) 1 c x = 2 c 3c2 (3c 2)(1 c) 3c 2 1c 2 x = lim or x = 5 c1 3c 2 y 1 3x 16 1 5 5 5 25 Therefore the centre of the required circle is 2 , 1 but circle passes through (2, 0) 5 25 2 2 2 1 2 64 1 1601 5 25 25 625 625 Radius of the required circle = 0 = Hence the required equation of the circle is x 22 y 1 2 1601 5 625 25 or 25x2 + 25y2 – 20x + 2y – 60 = 0 Ans. Illustration 24 : Two straight lines rotate about two fixed points. If they start from their position of coincidence such that one rotates at the rate double that of the other. Prove that the locus of their point of intersection is a circle. Solution : Let A (–a, 0) and B (a, 0) be two fixed points. Let one line which rotates about B an angle with the x-axis at any time t and at that time the Node6\\E_NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#05\\Eng\\02 CIRCLE.p65 second line which rotates about A make an angle 2 with x-axis. Now equation of line through B and A are respectively y – 0 = tan(x – a) ...... (i) and y – 0 = tan2(x + a) ...... (ii) From (ii), 2 tan 2 y (x a) B(a, 0) A(–a, 0) O(0, 0) 1 tan2 2y x a x a (from (i)) 1 y2 x a 2 2yx ax a y x a2 y2 (x – a)2 – y2 = 2(x2 – a2) or x2 + y2 + 2ax – 3a2 = 0 which is the required locus. 62 E
JEE-Mathematics Illustration 25 : If the circle x2 + y2 + 6x – 2y + k = 0 bisects the circumference of the circle x2 + y2 + 2x – 6y – 15 = 0, then k = (A) 21 (B) –21 (C) 23 (D) –23 Solution : 2g (g – g ) + 2f (f – f ) = c – c 21 2 2 1 2 12 2(1) (3 – 1) + 2 (–3) (–1 + 3) = k + 15 4 – 12 = k + 15 or –8 = k + 15 k = –23 Ans. (D) Illustration 26 : Find the equation of the circle of minimum radius which contains the three circles. S x2 + y2 – 4y – 5 = 0 1 S x2 + y2 + 12x + 4y + 31 = 0 2 S x2 + y2 + 6x + 12y + 36 = 0 3 Solution : For S1, centre = (0, 2) and radius = 3 For S , centre = (–6, –2) and radius = 3 (0,2) 2 P(a,b) For S , centre = (–3, –6) and radius = 3 3 let P(a, b) be the centre of the circle passing through the centres of the three given circles, then (a – 0)2 + (b – 2)2 = (a + 6)2 + (b + 2)2 (a + 6)2 – a2 = (b – 2)2 – (b + 2)2 (2a + 6)6 = 2b(–4) 2 6(a 3) 3 (a 3) b = 8 2 again (a – 0)2 + (b – 2)2 = (a + 3)2 + (b + 6)2 (a + 3)2 – a2 = (b – 2)2 – (b + 6)2 (2a + 3)3 = (2b + 4) (– 8) (2a + 3)3 = –16 3 (a 3 ) 2 2 6a + 9 = –8(–3a – 5) 6a + 9 = 24a + 40 18a = –31 a = 31 , b 23 18 12 radius of the required circle = 3 31 2 23 2 2 = 3 5 949 18 12 36 Node6\\E_NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#05\\Eng\\02 CIRCLE.p65 equation of the required circle is x 31 2 y 23 2 3 5 9 49 2 18 12 36 Illustration 27 : Find the equation of the image of the circle x2 + y2 + 16x – 24y + 183 = 0 by the line mirror 4x + 7y + 13 = 0. Solution : Centre of given circle = (–8, 12), radius = 5 the given line is 4x + 7y + 13 = 0 let the centre of required circle is (h, k) since radius will not change. so radius of required circle is 5. Now (h, k) is the reflection of centre (–8, 12) in the line 4x + 7y + 13 = 0 8 h , 12 k 2 2 Co-ordinates of A = (–8,12) A 4(8 h) 7(12 k ) 13 0 4x+7y+13=0 22 –32 + 4h + 84 + 7k + 26 = 0 4h + 7k + 78 = 0 .........(i) (h,k) E 63
JEE-Mathematics Also k 12 7 h 8 4 4k – 48 = 7h + 56 4k = 7h + 104 .........(ii) solving (i) & (ii) h = –16, k = –2 required circle is (x + 16)2 + (y + 2)2 = 52 Illustration 28 : The circle x2 + y2 – 6x – 10y + k = 0 does not touch or intersect the coordinate axes and the point (1, 4) is inside the circle. Find the range of the value of k. Solution : Since (1, 4) lies inside the circle S < 0 1 (1)2 + (4)2 – 6(1) – 10(4) + k < 0 k < 29 Also centre of given circle is (3, 5) and circle does not touch or intersect the coordinate axes r < CA & r < CB CA = 5 B r C(3,5) CB = 3 r r < 5 & r < 3 r < 3 or r2 < 9 r2 = 9 + 25 – k r2 = 34 – k 34 – k < 9 A k > 25 k (25, 29) Illustration 29 : The circle x2 + y2 – 4x – 8y + 16 = 0 rolls up the tangent to it at (2 + 3 , 3) by 2 units, find the equation of the circle in the new position. Solution : Given circle is x2 + y2 – 4x – 8y + 16 = 0 let P (2 + 3 , 3) B Equation of tangent to the circle at P(2 + 3 , 3) will be 2 Node6\\E_NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#05\\Eng\\02 CIRCLE.p65 (2 + 3 )x + 3y – 2(x + 2 + 3 ) – 4(y + 3) + 16 = 0 A or 3 x – y – 2 3 = 0 (2,4) slope = 3 tan = 3 P(2+ 3,3) = 60° line AB is parallel to the tangent at P coordinates of point B = (2 + 2cos60°, 4 + 2sin60°) thus B = (3, 4 + 3 ) radius of circle = 22 42 16 2 equation of required circle is (x – 3)2 + (y – 4 – 3 )2 = 22 Illustration 30 : A fixed circle is cut by a family of circles all of which, pass through two given points A(x , y ) and 11 B(x , y ). Prove that the chord of intersection of the fixed circle with any circle of the family passes 22 through a fixed point. Solution : Let S = 0 be the equation of fixed circle S=0 S1=0 A(x1,y1) let S = 0 be the equation of any circle through A and B 1 which intersect S = 0 in two points. L S – S = 0 is the equation of the chord of intersection 1 of S = 0 and S = 0 B 1 (x2,y2) L=0 let L = 0 be the equation of line AB 1 64 E
JEE-Mathematics let S be the equation of the circle whose diametrical ends are A(x , y ) & B(x , y ) 2 11 22 then S S – L1 = 0 1 2 L S – (S – L1) = 0 or L (S – S ) + L1 = 0 2 2 or L L' + L1 = 0 ........(i) (i) implies each chord of intersection passes through the fixed point, which is the point of intersection of lines L' = 0 & L = 0. Hence proved. 1 Illustration 31 : Let L1 be a straight line through the origin and L2 be the straight line x + y = 1. If the intercepts made by the circle x2 + y2 – x + 3y = 0 on L1 & L2 are equal , then which of the following equations can represent L1? (A) x + y = 0 (B) x – y = 0 (C) x + 7y = 0 (D) x – 7y = 0 Solution : Let L be y = mx 1 lines L & L will be at equal distances from centre of the circle centre of the circle is 1 , 3 1 2 2 2 1 m 3 1 3 1 (m 3)2 8 2 22 2 (1 m2 ) 1 m2 2 7m2 – 6m – 1 = 0 (m – 1) (7m + 1) = 0 1 y = x, 7y + x = 0 Ans. (B, C) m = 1, m = – 7 ANSWERS FOR DO YOURSELF 1 : (i) Centre 3 , 5 , Radius 3 10 ( i i ) 17(x2 + y2) + 2x – 44y = 0 4 4 4 (i i i ) x = p (1 2 cos ) ; y p (1 2 sin ) (iv) x2 + y2 + 6x – 2y – 51 = 0 22 2 : ( i ) (1, 2) lie inside the circle and the point (6, 0) lies outside the circle ( i i ) min = 0, max = 6, power = 0 Node6\\E_NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#05\\Eng\\02 CIRCLE.p65 3 : ( i ) xcos + ysin = a(1 + cos) ( i i ) 4x – 3y + 7 = 0 & 4x – 3y – 43 = 0 (iii) 5x + 12y = ±26 ; 10 , 24 (iv) 1 13 13 4 : ( i ) x + 2y = 1 5 : ( i ) 4x + 7y + 10 = 0 405 3 ( i i ) sq. units 6 : ( i ) 5x – 4y + 26 = 0 7 : ( i ) (x – h)2 + (y – k)2 = 2a2 52 ( i i ) x2 + y2 + gx + ƒy = 0 (ii) 10 (iii) angle between the tangents = 90° 8 : ( i i ) x2 y2 10x 10y 12 0 (i i i ) x2 + y2 + 4x – 7y + 5 = 0 7 77 9 : (i ) (x – 5)2 + (y – 5)2 = 25 (ii) 4 10 : (i) 135° ( i i ) x + 2y = 2 (i i i ) (1, 2) 11 : (i) 18 ( i i ) a–2 + b–2 = c–1 E 65
Search
Read the Text Version
- 1
- 2
- 3
- 4
- 5
- 6
- 7
- 8
- 9
- 10
- 11
- 12
- 13
- 14
- 15
- 16
- 17
- 18
- 19
- 20
- 21
- 22
- 23
- 24
- 25
- 26
- 27
- 28
- 29
- 30
- 31
- 32
- 33
- 34
- 35
- 36
- 37
- 38
- 39
- 40
- 41
- 42
- 43
- 44
- 45
- 46
- 47
- 48
- 49
- 50
- 51
- 52
- 53
- 54
- 55
- 56
- 57
- 58
- 59
- 60
- 61
- 62
- 63
- 64
- 65
- 66
- 67
- 68
- 69
- 70
- 71
- 72
- 73
- 74
- 75
- 76
- 77
- 78
- 79
- 80
- 81
- 82
- 83
- 84
- 85
- 86
- 87
- 88
- 89
- 90
- 91
- 92
- 93
- 94
- 95
- 96
- 97
- 98
- 99
- 100
- 101
- 102
- 103
- 104
- 105
- 106
- 107
- 108
- 109
- 110
- 111
- 112
- 113
- 114
- 115
- 116
- 117
- 118
- 119
- 120
- 121
- 122
- 123
- 124
- 125
- 126
- 127
- 128
- 129
- 130
- 131
- 132
- 133
- 134
- 135
- 136
- 137
- 138
- 139
- 140
- 141
- 142
- 143
- 144
- 145
- 146
- 147
- 148
- 149
- 150
- 151
- 152
- 153
- 154
- 155
- 156
- 157
- 158
- 159
- 160
- 161
- 162
- 163
- 164
- 165
- 166
- 167
- 168
- 169
- 170
- 171
- 172
- 173
- 174
- 175
- 176
- 177
- 178
- 179
- 180
- 181
- 182
- 183
- 184
- 185
- 186
- 187
- 188
- 189
- 190
- 191
- 192
- 193
- 194
- 195
- 196
- 197
- 198
- 199
- 200
- 201
- 202
- 203
- 204
- 205
- 206
- 207
- 208
- 209
- 210
- 211
- 212
- 213
- 214
- 215
- 216
- 217
- 218
- 219
- 220
- 221
- 222
- 223
- 224
- 225
- 226
- 227
- 228
- 229
- 230
- 231
- 232
- 233
- 234
- 235
- 236
- 237
- 238
- 239
- 240
- 241
- 242
- 243
- 244
- 245
- 246
- 247
- 248
- 249
- 250
- 251
- 252
- 253
- 254
- 255
- 256
- 257
- 258
- 259
- 260
- 261
- 262
- 263
- 264
- 265
- 266
- 267
- 268
- 269
- 270
- 271
- 272
- 273
- 274
- 275
- 276
- 277
- 278
- 279
- 280
- 281
- 282
- 283
- 284
- 285
- 286
- 287
- 288
- 289
- 290
- 291
- 292
- 293
- 294
- 295
- 296
- 297
- 298
- 299
- 300
- 301
- 302
- 303
- 304
- 305
- 306
- 307
- 308
- 309
- 310
- 311
- 312
- 313
- 314
- 315
- 316
- 317
- 318
- 319
- 320
- 321
- 322
- 323
- 324
- 325
- 326
- 327
- 328
- 329
- 330
- 331
- 332
- 333
- 334
- 335
- 336
- 337
- 338
- 339
- 340
- 341
- 342
- 343
- 344
- 345
- 346
- 347
- 348
- 349
- 350
- 351
- 352
- 353
- 354
- 355
- 356
- 357
- 358
- 359
- 360
- 361
- 362
- 363
- 364
- 365
- 366
- 367
