jmc

Let the coordinates of point $P$ be $(a,b)$. We have $a^2+b^2=10$ since $AP=\sqrt{10}$, and $a^2+(b-3{)}^2=13$ since $AB=\sqrt{13}$. Expanding $(b-3{)}^2$ gives us $$a^2+b^2-6b+9=13.$$Since $a^2+b^2=10$, we have $10-6b+9=13$, so $b=1$. From $a^2+b^2=10$, we have $a^2=9$, so $a=\pm 3$. If $a$ is $-3$, the point is not inside the triangle, so $a=3$. So the point is $(3,1)$ and the distance from $C$ is $$\sqrt{(3-5{)}^2+1^2}=\boxed{\sqrt{5}}.$$