jmc

We can draw another radius from the center to the point of tangency. This angle, $\angle ODB$, is $90^{\circ }$. Label the center $O$, the point of tangency $D$, and the radius $r$. Since $ODBC$ is a kite, then $DB=CB=5$. Also, $AD=13-5=8$. By the Pythagorean Theorem, $r^2+8^2=(12-r{)}^2$. Solving, $r^2+64=144-24r+r^2\Rightarrow 24r=80\Rightarrow \boxed{\frac{10}{3}}$.