Isosceles $\triangle ABC$ has a right angle at $C$. Point $P$ is inside $\triangle ABC$ such that $PA=9, PB=7, PC=4$. Legs $\overline{AC}$ and $\overline{BC}$ have length $s=\sqrt{a+b\sqrt2}$, where $a,b$ are positive integers. What is $a+b$?

1. Rotating $\triangle BCP$ $90^\circ$ to $\triangle ACP'$ gives $AP'=PB=7$, $PP'=PC\sqrt2=4\sqrt2$, $PP'^2=32$.
2. Check $PP'^2+AP'^2=32+49=81=PA^2$, so $\angle AP'P=90^\circ$ (theorem 19), giving $\angle AP'C=90^\circ+45^\circ=135^\circ$.
3. $AC^2=CP'^2+AP'^2+2\cdot CP'\cdot AP'\cdot\frac{\sqrt2}2=16+49+4\cdot7\sqrt2=65+28\sqrt2$.
4. $a+b=65+28=\boxed{93}$.