imc

We directly apply Euler's Theorem, which states that if the circumcenter is $O$ and the incenter $I$, and the inradius is $r$ and the circumradius is $R$, then $$O{I}^2=R(R-2r)$$ We can see that this is a right triangle, and hence has area $30$. We then find the inradius with the formula $A=rs$, where $s$ denotes semiperimeter. We easily see that $s=15$, so $r=2$. We now find the circumradius with the formula $A=\frac{abc}{4R}$. Solving for $R$ gives $R=\frac{13}{2}$. Additionally, we may notice that the side lengths $(5,12,13)$ are in a Pythagorean triple, and therefore the triangle is right for a circumradius of $\frac{13}{2}$. Substituting all of this back into our formula gives: $$O{I}^2=\frac{65}{4}$$ So, $OI=\frac{\sqrt{65}}{2}⟹\boxed{\frac{\sqrt{65}}{2}}$