Let C _ 1 be the circle of radius 1 with center at the origin. Let...

Plotting the diagram of the given value we get,

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

${C}_{1}: {x}^{2} + {y}^{2} = 1$

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

$⇒ {x}^{2} + {y}^{2} - 8x - 2y + 17 - {r}^{2} = 0$

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,

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

Also given, $AB = \sqrt{5}$

Now using distance formula we get,

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

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

$⇒ \dfrac{18 - {r}^{2}}{8} - 4 = \pm 2$

Taking positive sign we get,

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

Now taking negative sign we get,

$⇒ \dfrac{18 - {r}^{2}}{8} = 2 ⇒ {r}^{2} = 2$

Hence, ${r}^{2} = 2$