imc

Opposite angles of every cyclic quadrilateral are supplementary, so $$\angle B+\angle D=180^{\circ }.$$ We claim that $AC=25.$ We can prove it by contradiction: If $AC<25,$ then $\angle B$ and $\angle D$ are both acute angles. This arrives at a contradiction. If $AC>25,$ then $\angle B$ and $\angle D$ are both obtuse angles. This arrives at a contradiction. By the Inscribed Angle Theorem, we conclude that $\overline{AC}$ is the diameter of the circle. So, the radius of the circle is $r=\frac{AC}{2}=\frac{25}{2}.$ The area of the requested region is $$\pi r^2-\frac{1}{2}\cdot AB\cdot BC-\frac{1}{2}\cdot AD\cdot DC=\frac{625\pi }{4}-\frac{168}{2}-\frac{300}{2}=\frac{625\pi -936}{4}.$$ Therefore, the answer is $a+b+c=\boxed{1565}.$