All Common Tangents

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 $x^2 + 2xy + y^2 + \sqrt{2}x - \sqrt{2}y = 0$ and finding $\theta$ we obtain:

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

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

$\implies 2x'^2 - 2y'^2 = 0 \implies \boxed{y' = x'^2}$ .

In a similar manner replacing $x$ and $y$ in $x^2 + 2xy + y^2 - \sqrt{2}x + \sqrt{2}y + 8 = 0$ we obtain: $\boxed{y' = -x'^2 - 4}$

Let $f(x') = x'^2$ and $g(x') = -x'^2 - 4$ $\implies f'(a) = 2a = g'(b) = -2b \implies a = -b$ .

Let $R:(-b,b^2)$ and $S:(b,-b^2 - 4) \implies m_{RS} = \dfrac{2b^2 + 4}{-2b} = - 2b$

$\implies 2b^2 -4 = 0 \implies b = \pm\sqrt{2}$ .

In the $x'y'$ system:

For $b = \sqrt{2} \implies R:(-\sqrt{2},2)$ and $S:(\sqrt{2},-6)$

and

For $b = -\sqrt{2} \implies P:(\sqrt{2},2)$ and $Q:(-\sqrt{2},-6)$

For region $R_{1}:$ $\overleftrightarrow{\rm RP} = 2$ and $f(x') = x'^2$

$\implies A_{R_{1}} = \displaystyle\int_{-\sqrt{2}}^{\sqrt{2}} (2 - x'^2) dx =$ $(2x' - \dfrac{x'^3}{3})|_{-\sqrt{2}}^{\sqrt{2}} = \dfrac{8\sqrt{2}}{3}$ .

For region $R_{2}:$ $\overleftrightarrow{\rm QS} = -6$ and $f(x') = -x'^2 - 4$

$\implies A_{R_{1}} = \displaystyle\int_{-\sqrt{2}}^{\sqrt{2}} (2 - x'^2) dx = A_{R_{1}} = \dfrac{8\sqrt{2}}{3}$

$\implies A_{R_{1} + R_{2}} = \dfrac{16\sqrt{2}}{3} = \dfrac{a\sqrt{b}}{c} \implies a + b + c = \boxed{21}$