Let ABC be the triangle with AB = 1, AC = 3 and ∠ BAC = (π)/(2) ....

Let $A$ be the origin $B$ on $x -$axis, $C$ on $y -$axis,

So on plotting the diagram we get,

Now equation of circumcircle is given by

${\left(x - \dfrac{1}{2}\right)}^{2} + {\left(y - \dfrac{3}{2}\right)}^{2} = {\left(\dfrac{1}{2}\right)}^{2} + {\left(\dfrac{3}{2}\right)}^{2} = \dfrac{5}{2} ⋯ \left(1\right)$

Required equation of circle touches $AB$ and $AC$ and have radius $r$ so its equation is given by,

${\left(x - r\right)}^{2} + {\left(y - r\right)}^{2} = {r}^{2} ⋯ \left(2\right)$

Now given circle in equation \left(2\right)$touches circumcircle internally, so by property distance between the centre will be equal to modulus of difference of radii,</p><p>So,${d}_{{c}_{1}{c}_{2}} = \left|{r}_{1} - {r}_{2}\right|$</p><p>$⇒ {\left(\dfrac{1}{2} - r\right)}^{2} + {\left(\dfrac{3}{2} - r\right)}^{2} = {\left(\left|\sqrt{\dfrac{5}{2}} - r\right|\right)}^{2}$</p><p>$⇒ \dfrac{1}{4} + {r}^{2} - r + \dfrac{9}{4} + {r}^{2} - 3r = {\left(\sqrt{\dfrac{5}{2}} - r\right)}^{2}$</p><p>$⇒ 2{r}^{2} - 4r + \dfrac{5}{2} = \dfrac{5}{2} + {r}^{2} - \sqrt{10}r$</p><p>$⇒ r = 0$or$r = 4 - \sqrt{10} = 0 . 837$</p><p>So,$r = 0 . 84 (on rounding off)