jmc

Let $O=(0,0),$ $A=(3,4),$ $B=(6,8),$ and $C=(5,13).$ Let $T$ be a point on the circumcircle of triangle $ABC,$ so that $\overline{OT}$ is tangent to the circumcircle. Note that $O,$ $A,$ and $B$ are collinear.



Then by power of a point, $O{T}^2=OA\cdot OB=5\cdot 10=50,$ so $OT=\sqrt{50}=\boxed{5\sqrt{2}}.$