Circle Fiesta.

The desired area $A = A_{R_{2}} - A_{R_{1}}$ .

$R_{2}:$

$x^2 + (y - 5)^2 = 25$

$x^2 + y^2 = 100$

Finding points of intersection:

$(x - 5)^2 + (\sqrt{100 - x^2} - 10)^2 = 25 \implies 20 - x = 2\sqrt{100 - x^2} \implies$ $5x(x - 8) = 0 \implies x = 0, x = 8$

$y_{2}(x) = \sqrt{100 - x^2}$ and $y_{1}(x) = -\sqrt{25 - (x - 5)^2} + 10 \implies$

$A_{R_{2}} = \displaystyle\int_{0}^{8} y_{2}(x) - y_{1}(x) dx =$ $\displaystyle\int_{0}^{8} \sqrt{100 - x^2} + \sqrt{25 - (x - 5)^2} - 10 dx$

For $I_{1} = \displaystyle\int \sqrt{25 - (x - 5)^2} dx$

Let $x - 5 = 5\sin(\theta) \implies dx = 5\cos(\theta) d\theta \implies$

$I_{1} = \dfrac{25}{2} \displaystyle\int (1 + \cos(2\theta)) d\theta = \dfrac{25}{2}(\theta + \sin(\theta)\cos(\theta)) =$

$\dfrac{25}{2}(\arcsin(\dfrac{x - 5}{5}) + (\dfrac{x - 5}{5})\dfrac{\sqrt{25 - (x - 5)^2}}{5})$

$\implies I_{1}|_{0}^{8} - \displaystyle\int_{0}^{8} 10 dx = (\implies I_{1}|_{0}^{8} - 80 = \dfrac{25}{2}\arcsin(\dfrac{3}{5}) + \dfrac{25\pi}{4} - 74$ .

Let $I_{2} = \displaystyle\int \sqrt{100 - x^2} dx$

Let $x = 10\sin(\theta) \implies dx = 10\cos(\theta) d\theta \implies$

$I_{2} = 50\displaystyle\int (1 + \cos(2\theta)) d\theta = 50(\theta + \sin(\theta)\cos(\theta)) =$

$50(\arcsin(\dfrac{x}{10}) + (\dfrac{x}{10})\dfrac{\sqrt{100 - x^2}}{10}) \implies$

$I_{2}|_{0}^{8} = 50\arcsin(\dfrac{4}{5}) + 24$

$\implies A_{R_{2}} = \boxed{\dfrac{25}{2}\arcsin(\dfrac{3}{5}) + 50\arcsin(\dfrac{4}{5}) + \dfrac{25\pi}{4} - 50}$ .

$R_{1}:$

$x^2 + (y - 5)^2 = 25$

$(x - 5)^2 + (y - 10)^2 = 25$

Finding points of intersection:

$y = \sqrt{25 - x^2} + 5 \implies (x - 5)^2 + ( \sqrt{25 - x^2} - 5)^2 = 25 \implies$

$\implies 5 - x = \sqrt{25 - x^2} \implies x(x - 5) = 0 \implies x = 0, 5$

$y_{2}(x) = \dfrac{25 - x^2} + 5$ and $y_{1}(x) = -\sqrt{25 - (x - 5)^2} + 10 \implies$

$A_{R_{1}} = \displaystyle\int_{0}^{5} y_{2}(x) - y_{1}(x) dx =$ $\displaystyle\int_{0}^{5} \sqrt{25 - x^2} + \sqrt{25 - (x - 5)^2} - 5 dx$

$I_{1} = \displaystyle\int_{0}^{5} \sqrt{25 - (x - 5)^2} - 5 dx =$

$\dfrac{25}{2}(\arcsin(\dfrac{x - 5}{5}) + (\dfrac{x - 5}{5})\dfrac{\sqrt{25 - (x - 5)^2}}{5})|_{0}^{5} - 25 = \dfrac{25}{4}\pi - 25$

For $I_{2} = \displaystyle\int_{0}^{5} \sqrt{25 - x^2} dx$

Let $x = 5\sin(\theta) \implies dx = 5\cos(\theta) \implies$

$I_{2} = \dfrac{25}{2}\displaystyle\int_{0}^{\dfrac{\pi}{2}} (1 + \cos(2\theta)) d\theta =$

$\dfrac{25}{2}(\theta + \sin(\theta)\cos(\theta))|_{0}^{\dfrac{\pi}{2}} = \dfrac{25\pi }{4}$

$\implies A_{R_{1}} = \boxed{\dfrac{50\pi}{4} - 25}$

$\implies$ The desired area $A = \dfrac{25}{2}\arcsin(\dfrac{3}{5}) + 50\arcsin(\dfrac{4}{5}) - \dfrac{25\pi}{4} - 25 =$

$\dfrac{5^2}{2}\arcsin(\dfrac{3}{5}) + 5^2 * 2\arcsin(\dfrac{2^2}{5}) - \dfrac{5^2}{2^2}\pi - 5^2 =$

$\dfrac{a^b}{b}\arcsin(\dfrac{c}{a}) + a^b * b\arcsin(\dfrac{b^b}{a}) - \dfrac{a^b}{b^b}\pi - a^b$

$\implies a + b + c = \boxed{10}$ .