Let $C_{1}$ be the circle of radius 1 with centre at the origin. Let $C_{2}$ be the circle of radius r with centre at the point A(6, 1), where 1 < r < 4. Two distinct common tangents PQ and ST of $C_{1}$ and $C_{2}$ are drawn. The tangent PQ touches $C_{1}$ at P and $C_{2}$ at Q. The tangent ST touches $C_{1}$ at S and $C_{2}$ at T. Midpoints of the line segment PQ and ST are joined to form a line which meets the x–axis at point B. If $A B = \sqrt{10}$ , then the value
of $r^{2}$ is _______

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answer is 2.

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Detailed Solution

$C_{1} : x^{2} + y^{2} = 1 C_{2} : {(x - 6)}^{2} + {(y - 1)}^{2} = r^{2}$
M is point of line segment PQ, N is mid point of line segment ST.
$\Rightarrow$ MN is radical axis of the two circles $C_{1}$ and $C_{2}$
$12 x + 2 y - 38 + r^{2} = 0$
$B (\frac{38 - r^{2}}{12}, 0)$
$A B = \sqrt{10}$ $\Rightarrow {(\frac{38 - r^{2}}{12} - 6)}^{2} + {(0 - 1)}^{2} = 10$
$\Rightarrow {(\frac{r^{2} + 34}{12})}^{2} = 9 \Rightarrow r^{2} + 34 = 36 \Rightarrow r^{2} = 2$