8. Given the vertices of $\triangle O A B$ are $O(0,0), A(4,4 \sqrt{3}), B(8,0)$, and its incenter is $I$. Let circle $C$ pass through points $A$ and $B$, and intersect circle $I$ at points $P$ and $Q$. If the tangents to the two circles at points $P$ and $Q$ are perpendicular, then the radius of circle $C$ is $\qquad$ .
Given that $C$ is on $O I: y=\frac{\sqrt{3}}{3} x$, let $C\left(t, \frac{\sqrt{3} t}{3}\right)$
$\bigodot I:(x-4)^{2}+\left(y-\frac{4 \sqrt{3}}{3}\right)^{2}=\frac{16}{3}, \bigodot C:$
$$
(x-t)^{2}+\left(y-\frac{\sqrt{3} t}{3}\right)^{2}=(t-8)^{2}+\left(\frac{\sqrt{3} t}{3}\right)^{2} \text {, }
$$

Subtracting the two circle equations, we get $P Q: x+\frac{\sqrt{3}}{3} y-8=\frac{8}{4-t}$.
Solving the system $\left\{\begin{array}{l}x+\frac{\sqrt{3}}{3} y-8=\frac{8}{4-t}, \\ y=\frac{\sqrt{3}}{3} x\end{array} \Rightarrow x_{D}=\frac{6}{4-t}+6\right.$, by the projection theorem we have $I P^{2}=I D \cdot I C$ $\Rightarrow \frac{16}{3}=\frac{4}{3}\left(\frac{6}{4-t}+6-4\right)(t-4) \Rightarrow t=9$, thus $C A^{2}=1+27=28$. Therefore, the radius of circle $C$ is $2 \sqrt{7}$.