Given, $\tan \alpha=\frac{1}{8}, \tan \beta=\frac{1}{2}$,
then $\tan (\alpha+\beta)=\frac{\frac{1}{8}+\frac{1}{2}}{1-\frac{1}{16}}=\frac{2}{3}$.
Thus, $\tan 2(\alpha+\beta)=\frac{2 \cdot \frac{2}{3}}{1-\frac{4}{9}}=\frac{12}{5}$
$\Rightarrow \sin \angle A P B=\frac{12}{13} \Rightarrow 2 R=\frac{10}{\frac{12}{13}}=\frac{65}{6} \Rightarrow R=\frac{65}{12}$
6. For the hyperbola $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{16}=1$, the left and right foci are $A, B$, and $P$ is a point on the hyperbola. If the incenter of $\triangle P A B$ is $(3,1)$, then the radius of the circumcircle of $\triangle P A B$ is $\qquad$ .