There's Circles Everywhere!

$C_{1}: (x - 20)^2 + y^2 = 400$

$C_{2}: x^2 + y^2 =400$

$C_{3}: (x - 20)^2 + (y - 20)^2 = 400$

$C_{4}: x^2 + (y - 20)^2 = 400$

We want the intersection of $C_{1}$ and $C_{2}$ , $C_{3}$ and $C_{4}$ and $C_{1}$ and $C_{3}$ .

Expanding $C_{1}$ and adding $C_{1}$ and $C_{2}$ we obtain $P_{C1C2} : (10,10\sqrt{3})$

Expanding $C_{3}$ and $C_{4}$ and adding we obtain $P_{C3C4} : (10,-10\sqrt{3} + 20)$

Expanding $C_{1}$ and $C_{3}$ and adding we obtain $P_{C1C3} : (-10\sqrt{3} + 20),10$

For $C_{1}$ we have $y_{2} = \sqrt{400 - (x - 20)^2}$

For $C_{3}$ we have $y_{1} = 20 - \sqrt{400 - (x - 20)^2}$

$\implies A = 2 * R_{1} = 2\displaystyle\int_{-10\sqrt{3} + 20}^{10} (2\sqrt{400 - (x - 20)^2} - 20) dx$

Letting $x - 20 = 20\sin(\theta) \implies dx = 20\cos(\theta) \implies$

$A = 2(\displaystyle\int_{-\dfrac{\pi}{3}}^{-\dfrac{\pi}{6}} (400(1 + \cos(2\theta)) d\theta -$ $20x|_{-10\sqrt{3} + 20}^{10}) =$

$2(400(\theta + \dfrac{1}{2}\sin(2\theta))|_{-\frac{\pi}{3}}^{-\frac{\pi}{6}}) - 20x|_{-10\sqrt{3} + 20}^{10})$

$= 2(\dfrac{200\pi}{3} + 200 -200\sqrt{3}) = 400(1 + \dfrac{\pi}{3} - \sqrt{3}) =$

$(5 * 2^2)^2( 1 + \dfrac{\pi}{3} - \sqrt{3}) = (a * b^b)^b(c + \dfrac{\pi}{d} - \sqrt{d}) \implies$ $a + b + c + d = \boxed{11}$ .