jmc

Let $X$ and $Y$ be the feet of the perpendiculars from $P$ and $Q$ to $BC$, respectively. Let the radius of $\odot Q$ be $r$. We know that $PQ=r+16$. From $Q$ draw segment $\overline{QM}\parallel \overline{BC}$ such that $M$ is on $PX$. Clearly, $QM=XY$ and $PM=16-r$. Also, we know $QPM$ is a right triangle. To find $XC$, consider the right triangle $PCX$. Since $\odot P$ is tangent to $\overline{AC},\overline{BC}$, then $PC$ bisects $\angle ACB$. Let $\angle ACB=2\theta$; then $\angle PCX=\angle QBX=\theta$. Dropping the altitude from $A$ to $BC$, we recognize the $7-24-25$ right triangle, except scaled by $4$. So we get that $\tan (2\theta )=24/7$. From the half-angle identity, we find that $\tan (\theta )=\frac{3}{4}$. Therefore, $XC=\frac{64}{3}$. By similar reasoning in triangle $QBY$, we see that $BY=\frac{4r}{3}$. We conclude that $XY=56-\frac{4r+64}{3}=\frac{104-4r}{3}$. So our right triangle $QPM$ has sides $r+16$, $r-16$, and $\frac{104-4r}{3}$. By the Pythagorean Theorem, simplification, and the quadratic formula, we can get $r=44-6\sqrt{35}$, for a final answer of $\boxed{254}$.