look here and here

Trig-able. Let $M,N$ be the midpoints of $BI$ and $CI$. Notice that we just need to prove that $O$ has the same power w.r.t. c and d. We can calculate the segment $OM$ since it is the median of triangle $IOB$ and we know the lengths $OI$ and $BI$ ($OI^2=R^2-2Rr$ and $BI=\frac{r}{sin(\frac{\alpha}{2})}$) and we're done by bashing it out.

Let $B'$ be the image of $B$ in $IC$. Define $C'$ similarly. Notice that the center of the image of the circle with diameter $IB$ in $IC$ is nothing but $M$ the midpoint of $IB'$, and similarly $N$ the midpoint of $IC'$ is the center of the other circle. Since $I$ is in both image circles, showing that $O$ lies on their radical axis is equivalent to showing that $OI \perp MN$, and since $MN$ is midline in $\triangle IB'C'$, this is equivalent to showing that $OI \perp B'C'$, or $OB'^2-OC'^2=IB'^2-IC'^2=IB^2-IC^2$, or $(OB'^2-R^2)-(OC'^2-R^2)=(IB^2-r^2)-(IC^2-r^2)$ or $B'A\cdot B'C-C'A\cdot C'B=BD^2-CD^2$, but this is just simple and trivial calculations.