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 $PQ$ 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 segments $PQ$ and $ST$ are joined to form a line which meets the $x$-axis at a point $B$. If $AB=\sqrt{5}$, then the value of $r^2$ is

Plotting the diagram of the given value we get,

Now let $M$ and $N$ be midpoints of $PQ$ and $ST$ respectively.
$\Rightarrow MN$ is a radical axis of two circles

$C_1:x^2+y^2=1$

$C_2:(x-4{)}^2+(y-1{)}^2=r^2$

$\Rightarrow x^2+y^2-8x-2y+17-r^2=0$

Now subtracting equation of both above circles we get,

Equation of $MN: 8x+2y-18+r^2=0$

Now the line $MN$ intersect the $x-\text{axis}$ at $B$

So, $B$ will be,

$\Rightarrow B\left(\frac{18-r^2}{8},0\right)$

Also given, $AB=\sqrt{5}$

Now using distance formula we get,

$\sqrt{{\left(\frac{18-r^2}{8}-4\right)}^2+1}=\sqrt{5}$

$\Rightarrow {\left(\frac{18-r^2}{8}-4\right)}^2=4$

$\Rightarrow \frac{18-r^2}{8}-4=\pm 2$

Taking positive sign we get,

$\Rightarrow \frac{18-r^2}{8}=6\Rightarrow r^2=-30 \left(\text{rejected}\right)$

Now taking negative sign we get,

$\Rightarrow \frac{18-r^2}{8}=2\Rightarrow r^2=2$

Hence,  $r^2=2$