Quadrilateral $ABCD$ with side lengths $AB=33, BC=56, CD=39, DA=52$ is inscribed in a circle. The area interior to the circle but exterior to the quadrilateral can be written as $\frac{a\pi-b}c$, where $a,b,c$ are positive integers with $\gcd(a,c)=1$. What is $a+b+c$?

1. Since $ABCD$ is cyclic, $\cos D=-\cos B$. The law of cosines for $AC$ in $\triangle ABC$ and $\triangle ADC$ gives $33^2+56^2-2\cdot33\cdot56\cos B=39^2+52^2+2\cdot39\cdot52\cos B$. As $33^2+56^2=4225=39^2+52^2$, this forces $\cos B=0$. So $\angle B=\angle D=90^\circ$ and $AC=65$ is a diameter.
2. $[ABCD]=\frac12(33)(56)+\frac12(39)(52)=924+1014=1938$.
3. Area $=\pi\left(\frac{65}2\right)^2-1938=\frac{4225\pi-7752}4$; since $\gcd(4225,4)=1$, $a+b+c=4225+7752+4=\boxed{11981}$.