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JEE Advanced Mathematics Circle 2023 JEE Advanced 2023 (Paper 2)

JEE Advanced Mathematics Question (2023) — Solution

Question

Let C 1  be the circle of radius 1 with center at the origin. Let C 2 be the circle of radius r with center at the point A = ( 4 , 1 ) , where 1 < r < 3 . Two distinct common tangents P Q and S T of C 1 and C 2 are drawn. The tangent P Q touches C 1 at P and C 2 at Q . The tangent S T touches C 1 at S and C 2 at T . Midpoints of the line segments P Q and S T are joined to form a line which meets the x -axis at a point B . If A B = 5 , then the value of r 2 is

Options

  1. A. A
  2. B. B
  3. C. C
  4. D. D

Answer

A. A

Step-by-step solution

Plotting the diagram of the given value we get, Now let M  and N be midpoints of P Q and S T respectively. ⇒ M N is a radical axis of two circles C 1 : x 2 + y 2 = 1 C 2 : ( x - 4 ) 2 + ( y - 1 ) 2 = r 2 ⇒ x 2 + y 2 - 8 x - 2 y + 17 - r 2 = 0 Now subtracting equation of both above circles we get, Equation of M N :   8 x + 2 y - 18 + r 2 = 0 Now the line  M N  intersect the  x - axis  at  B So,  B  will be, ⇒ B 18 - r 2 8 , 0 Also given,  A B = 5 Now using distance formula we get, 18 - r 2 8 - 4 2 + 1 = 5 ⇒ 18 - r 2 8 - 4 2 = 4 ⇒ 18 - r 2 8 - 4 = ± 2 Taking positive sign we get, ⇒ 18 - r 2 8 = 6 ⇒ r 2 = - 30   rejected Now taking negative sign we get, ⇒ 18 - r 2 8 = 2 ⇒ r 2 = 2 Hence,   r 2 = 2

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