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
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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