Solution:

Let $\Gamma$ be the circumcircle of triangle $ABC$, and let $E$ be the center of the circle tangent to $\Gamma$ and the sides $AC$ and $BC$. Notice that $\angle C = 90^{\circ}$ because $33^{2} + 56^{2} = 65^{2}$. Let $D$ be the second intersection of line $CE$ with $\Gamma$, so that $D$ is the midpoint of the arc $AB$ away from $C$. Because $\angle BCD = 45^{\circ}$, one can easily calculate $CD = 89 \sqrt{2} / 2$. The power of $E$ with respect to $\Gamma$ is both $r(65 - r)$ and $r \sqrt{2} \cdot (89 \sqrt{2} / 2 - r \sqrt{2}) = r(89 - 2r)$, so $r = 89 - 65 = 24$.