Solution:

Let $S$, $T$ be the intersections of the tangents to the circumcircle of $ABC$ at $A$, $C$ and at $A$, $B$ respectively. Note that $ASCO$ is cyclic with diameter $SO$, so the circumcenter of $AOC$ is the midpoint of $OS$, and similarly for the other side. So the length we want is $\frac{1}{2} ST$. The circumradius $R$ of $ABC$ can be computed by Heron's formula and $K = \frac{abc}{4R}$, giving $R = \frac{65}{8}$. A few applications of the Pythagorean theorem and similar triangles gives $AT = \frac{65}{6}$, $AS = \frac{39}{2}$, so the answer is $\frac{91}{6}$.