Place $A$ at the origin with $B=(1,0)$ on the $x$-axis and $C=(0,3)$ on the $y$-axis. Since $\angle A=\frac{\pi }{2}$, the hypotenuse $BC$ is a diameter of the circumcircle.

The circumcentre is the midpoint of $BC$: $\left(\frac{1}{2},\frac{3}{2}\right)$, and the circumradius is $R=\frac{1}{2}\sqrt{1^2+3^2}=\frac{\sqrt{10}}{2}$.

A circle tangent to both axes lying in the first quadrant has centre $(r,r)$ and radius $r$. For internal tangency with the circumcircle, the distance between centres equals $R-r$:

$\sqrt{{\left(\frac{1}{2}-r\right)}^2+{\left(\frac{3}{2}-r\right)}^2}=\frac{\sqrt{10}}{2}-r.$

Squaring: $\frac{1}{4}+\frac{9}{4}-4r+2r^2=\frac{10}{4}-\sqrt{10}r+r^2$, which simplifies to $r^2-4r+\sqrt{10}r=0$, i.e., $r(r-4+\sqrt{10})=0$.

Since $r>0$, $r=4-\sqrt{10}\approx 0.8377\approx 0.84$ (rounded to two decimals).