Let $ABC$ be the triangle with $AB=1,AC=3$ and $\angle BAC=\frac{\pi }{2}$. If a circle of radius $r>0$ touches the sides $AB,AC$ and also touches internally the circumcircle of the triangle $ABC$, then the value of $r$ is ______. (round-off the value to two decimal place)

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 

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

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

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

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,

So, $d_{c_1{c}_2}=\left|r_1-r_2\right|$

$\Rightarrow {\left(\frac{1}{2}-r\right)}^2+{\left(\frac{3}{2}-r\right)}^2={\left(\left|\sqrt{\frac{5}{2}}-r\right|\right)}^2$

$\Rightarrow \frac{1}{4}+r^2-r+\frac{9}{4}+r^2-3r={\left(\sqrt{\frac{5}{2}}-r\right)}^2$

$\Rightarrow 2r^2-4r+\frac{5}{2}=\frac{5}{2}+r^2-\sqrt{10}r$

$\Rightarrow r=0$ or $r=4-\sqrt{10}=0.837$

So, $r=0.84$  (on rounding off)