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

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

TestHub · Accepted Answer

Plotting the diagram of the given value we get,
Now let $M$ and $N$ be midpoints of $P Q$ and $S T$ respectively.
$\Rightarrow M N$ is a radical axis of two circles
$C_{1} : x^{2} + y^{2} = 1$
$C_{2} : (x - 4)^{2} + (y - 1)^{2} = r^{2}$
$\Rightarrow x^{2} + y^{2} - 8 x - 2 y + 17 - r^{2} = 0$
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,
$\Rightarrow B (\frac{18 - r^{2}}{8}, 0)$
Also given, $A B = \sqrt{5}$
Now using distance formula we get,
$\sqrt{{(\frac{18 - r^{2}}{8} - 4)}^{2} + 1} = \sqrt{5}$
$\Rightarrow {(\frac{18 - r^{2}}{8} - 4)}^{2} = 4$
$\Rightarrow \frac{18 - r^{2}}{8} - 4 = \pm 2$
Taking positive sign we get,
$\Rightarrow \frac{18 - r^{2}}{8} = 6 \Rightarrow r^{2} = - 30 (rejected)$
Now taking negative sign we get,
$\Rightarrow \frac{18 - r^{2}}{8} = 2 \Rightarrow r^{2} = 2$
Hence, $r^{2} = 2$

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