It's All About Tangents.

$\textbf{To obtain graph below by using rotation equations to find angle}$ $\theta$ :

Using the equations of rotation for a rotation about the origin $(0,0):$

$x = x'\cos(\theta) - y'\sin(\theta)$

$y = x'\sin(\theta) + y'\cos(\theta)$

Replacing $x$ and $y$ in $x^2 + 2xy + y^2 + 2x - 2y - 3 = 0$ and finding $\theta$ we obtain:

$x'^2 + y'^2 + (x'^2 - y'2)\sin(2\theta) + 2\cos(2\theta)x'y' + 2x'(\cos(\theta) - \sin(\theta)) -2y'(\sin(\theta) + \cos(\theta)) - 3 = 0$

Setting $x'y'$ term to zero we have $\cos(2\theta) = 0 \implies 2\theta = \dfrac{\pi}{2} \implies \theta = \dfrac{\pi}{4}$ .

$\implies 2x'^2 - 2\sqrt{2}y - 3 = 0 \implies \boxed{y' = \dfrac{1}{2}(x'^2 - \dfrac{3}{2})}$ .

Since any circle is invariant under a rotation about its center we can write: $x'^2 + y'^2 = 1$

Solving the system $x'^2 + y'^2 = 1$ and $y' = \dfrac{1}{2}(x'^2 - \dfrac{3}{2})$ :

$x'^2 = \sqrt{2}y' + \dfrac{3}{2} \implies 2y'^2 + 2\sqrt{2}y' + 1 = 0 \implies y' = -\dfrac{1}{\sqrt{2}} \implies x' = \pm\dfrac{1}{\sqrt{2}}$

Using the bottom portion $y_{2}(x) = -\sqrt{1 - x'^2}$ of the circle and $y_{1}(x) = \dfrac{1}{2}(x'^2 - \dfrac{3}{2})$ . the area $A =$ $2\displaystyle\int_{0}^{\dfrac{1}{\sqrt{2}}} y_{2}(x) - y_{1}(x) dx$ .

Letting $x' = \sin(\theta) \implies dx' = \cos(\theta) d\theta \implies$

$A = 2(-\dfrac{1}{2}\displaystyle\int_{0}^{\dfrac{\pi}{4}} (1 + \cos(2\theta)) d\theta - \dfrac{1}{\sqrt{2}} (\dfrac{x'^3}{3} - \dfrac{3}{2}x')|_{0}^{\frac{1}{\sqrt{2}}})$

$= 2(-\dfrac{1}{2}(\theta + \dfrac{1}{2}\sin(2\theta))|_{0}^{\frac{\pi}{4}} + \dfrac{2}{3})$

$= \dfrac{5}{6} - \dfrac{\pi}{4} = \dfrac{5}{2 * 3} - \dfrac{\pi}{2^2} = \dfrac{c}{a * b} - \dfrac{\pi}{a^a} \implies a + b + c = \boxed{10}$ .