XV OBM

Let $H$ on $BC$ such that $AH=h$. Since $BAC$ is a right angle, $BH\cdot CH=h^2$.

But the power of $H$ with respect to the given circle is $HB\cdot HC=R^2-O{H}^2$.

So $OH=\sqrt{R^2-h^2}$ is determined and $H$ is the intersection of the circle with center $O$ and radius $\sqrt{R^2-h^2}$ and the circle with center $A$ and radius $h$. So $H$ is determined. To determine $B$ and $C$, it is enough to trace a perpendicular to $AH$ passing through $H$.