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 ________.

$AB=1, \quad AC=3, \quad \angle BAC=\dfrac{\pi}{2}=90^{\circ}$ $\text{circle of radius } r \text{ touches } AB,\ AC \text{ and the circumcircle}$ $\text{Find: } r = \,?$ $\text{circle tangent to } AB,\ AC \text{ and internally to the circumcircle}$ $A=(0,0), \quad B=(1,0), \quad C=(0,3)$ $BC=\sqrt{1^{2}+3^{2}}=\sqrt{10}$ $R=\dfrac{\sqrt{10}}{2}, \quad \text{centre}=\left(\dfrac{1}{2},\dfrac{3}{2}\right)$ $\text{circle touches } AB\ (y=0) \text{ and } AC\ (x=0)$ $\Rightarrow \text{centre}=(r,r)$ $1.\ \text{circumcentre } \left(\tfrac{1}{2},\tfrac{3}{2}\right),\ R=\tfrac{\sqrt{10}}{2}$ $2.\ \text{small circle centre } (r,r)$ $3.\ \text{internal tangency: } d=R-r \Rightarrow r$ $d = R - r$ $d = \sqrt{\left(r-\dfrac{1}{2}\right)^{2}+\left(r-\dfrac{3}{2}\right)^{2}}$ $\sqrt{\left(r-\dfrac{1}{2}\right)^{2}+\left(r-\dfrac{3}{2}\right)^{2}} = \dfrac{\sqrt{10}}{2}-r$ $\left(r-\dfrac{1}{2}\right)^{2}+\left(r-\dfrac{3}{2}\right)^{2} = 2r^{2}-4r+\dfrac{5}{2}$ $\left(\dfrac{\sqrt{10}}{2}-r\right)^{2} = r^{2}-\sqrt{10}\,r+\dfrac{5}{2}$ $2r^{2}-4r = r^{2}-\sqrt{10}\,r$ $r^{2}+(\sqrt{10}-4)r=0$ $r\left(r+\sqrt{10}-4\right)=0$ $r=0 \ \text{(discard)} \quad \text{or} \quad r=4-\sqrt{10}$ $r=4-\sqrt{10}\approx 0.84$ $d = R-r \approx 1.58-0.84 = 0.74 \ \checkmark$ $\boxed{r = 4-\sqrt{10}}$