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QJEEAdv23Ma963e: JEE Advanced 2023 Mathematics - Class 11 JEEAdv23Ma963e Solution: Let be the circle of radius 1 with center at the origin. Let be …

Let C1C_1 be the circle of radius 1 with center at the origin. Let C2C_2 be the circle of radius rr with center at the point A=(4,1)A=(4,1) , where 1 < r < 3 . Two distinct common tangents PQP Q and STS T of C1C_1 and C2C_2 are drawn. The tangent PQP Q touches C1C_1 at PP and C2C_2 at QQ . The tangent STS T touches C1C_1 at SS and C2C_2 at TT . Mid points of the line segments PQP Q and STS T are joined to form a line which meets the xx -axis at a point BB . If AB=5A B=\sqrt{5} , then the value of r2r^2 is :

Step-by-Step Verified Solution

JEE Advanced 2023 Paper 2 Online Mathematics - Circle Question 2 English Explanation
Let MM and NN be midpoints of PQP Q and STS T respectively.

⇒MN\Rightarrow M N is a radical axis of two circles

C1:x2+y2=1C_1: x^2+y^2=1 ........(i)

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From (i) and (ii);

Equation of MN:8x+2y−18+r2=0M N: 8 x+2 y-18+r^2=0

⇒B\Rightarrow B is on xx -axis ⇒B(18−r28,0)\Rightarrow B\left(\frac{18-r^2}{8}, 0\right)

AB=5(18−r28−4)2+1=5

⇒\Rightarrow On solving r2=2r^2=2

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