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JEE Advanced Mathematics Circle 2019 JEE Advanced 2019 (Paper 2)

JEE Advanced Mathematics Question (2019) — Solution

Question

Let the circles C 1 : x 2 + y 2 = 9 and C 2 : x - 3 2 + y - 4 2 = 16 , intersect at the points X and Y . Suppose that another circle C 3 : x - h 2 + y - k 2 = r 2 satisfies the following conditions: i centre of C 3 is collinear with the centres of C 1 and C 2 i i   C 1 and C 2 both lie inside C 3 , and i i i   C 3 touches C 1 at M and C 2 at N . Let the line through X and Y intersect C 3 at Z and W , and let a common tangent of C 1 and C 3 be a tangent to the parabola x 2 = 8 α y . There are some expression given in the List- I whose values are given in List- I I below:     List- I   List-  II I 2 h + k P 6 I I L e n g t h   o f   Z W L e n g t h   o f   X Y Q 6 I I I A r e a   o f   t r i a n g l e   M Z N A r e a   o f   t r i a n g l e   Z M W R 5 4 I V α S 21 5     T 2 6     U 10 3 Which of the following is the only INCORRECT combination?

Options

  1. A. I I - T
  2. B. I - S
  3. C. I - U
  4. D. I I - Q

Answer

D. I I - Q

Step-by-step solution

Given centre of C 1 , C 2 and C 3 are collinear hence 0 0 1 3 4 1 h k 1 = 0 3 k = 4 h …(i) ⇒ M N is diameter of C 3 M N = M C 1 + C 1 C 2 + C 2 N 2 r = r 1 + C 1 C 2 + r 2 ∵ M N = 2 r 2 r = 3 + 3 - 0 2 + 4 - 0 2 + 4 r = 6 …(ii) ⇒ Given C 3 touches C 1 at M So C 1 C 3 = r - 3 h 2 + k 2 = 9 …(iii) From (i) and (iii) h = ± 9 5 and k = ± 12 5 So centre of C 3 will be 9 5 , 12 5 ⇒ Now equation of common chord of C 1 and C 2 will be C 1 - C 2 = 0 6 x + 8 y = 18 Equation of line X Y is 3 x + 4 y = 9 …(iv) Distance of line X Y from origin C 1 P = 9 5 ⇒ now in ∆ C 1 P y C 1 P 2 + P Y 2 = C 1 Y 2 81 25 + P Y 2 = 9 ∵ C 1 Y = r 1 = 3 P Y 2 = 144 25 ⇒ P Y = 12 5 Length of X Y = 2 P Y = 24 5 ⇒ Line Z W is line X Y Equation of Z W = 3 x + 4 y = 9 Distance of C 3 from Z W = 3 9 5 + 4 12 5 - 9 5 Z W = 6 5 now Z W = 2 6 2 - 6 5 2 Z W = 24 6 5 I 2 h + k = 2 × 9 5 + 12 5 = 30 5 = 6 I I L e n g t h o f Z W L e n g t h o f X Y = 6 I I I A r e a o f ∆ M Z N A r e a o f ∆ Z M W = 1 2 × M N × P Z 1 2 × Z W × M P = 1 2 M N 1 2 Z W 1 2 × Z W M G + G P ∴ P Z = 1 2 Z W = 1 2 × 12 × 12 6 5 1 2 × 24 8 5 24 5 ∴ M C 1 = 3 ∴ C 1 A = 9 5 = 5 4 I V Common tangent to C 1 and C 3 is common chord to C 1 and C 3 C 1 - C 3 = 0 3 x + 4 y + 15 = 0 Now this line is tangent to parabola x 2 = 8 α y x 2 = 8 α - 3 x - 15 y 4 x 2 + 24 α x + 120 α = 0 Apply D = 0 (for tangent, it will have repeated roots) α = 10 3

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