Circle inscribed in two intersecting parabolas

$(1): \:\ x^2 + 2xy + y^2 - \sqrt{2}x + \sqrt{2}y = 8$

$(2): \:\ x^2 + 2xy + y^2 + \sqrt{2}x - \sqrt{2}y = 8$ .

The equations of rotation are:

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

Since $Coeff(x^2) = Coeff(y^2)$ in both curves $\implies \theta = 45^{\circ} \implies$

$x = \dfrac{x' - y'}{\sqrt{2}}$ and $y = \dfrac{x' + y'}{\sqrt{2}}$

$(1) \implies \dfrac{x'^2 - 2x'y' + y'^2}{2} + (x'^2 - y'^2) + \dfrac{x'^2 + 2x'y' + y'^2}{2} - (x' - y') + x' + y' = 8$

$\implies x'^2 + y' = 4 \implies \boxed{y' = 4 - x'^2}$ and similarly for $(2)$ we have: $\boxed{y' = x'^2 - 4}$ .

Using $y' = 4 - x'^2 \implies D = d^2 = x'^2 + (4 - x'^2)^2 = x'^4 - 7x'^2 + 16 \implies$

$\dfrac{dD}{dx'} = 2x'(2x'^2 - 7) = 0$ and $x' \neq 0 \implies x' = \pm\sqrt{\dfrac{7}{2}} \implies y' = \dfrac{1}{2} \implies$

$D = d^2 = \dfrac{15}{4}$

and $\dfrac{d^2D}{dx'^2} > 0 \implies$ the distance $d$ is minimized when $x' = \pm\sqrt{\dfrac{7}{2}}$

For $R_{1}:$

$x'^2 + y'^2 = \dfrac{15}{4} \implies y' = \sqrt{\dfrac{15}{4} - x'^2}$ and $y' = 4 - x'^2 \implies$

$A = \displaystyle\int_{-\sqrt{\frac{7}{2}}}^{\sqrt{\frac{7}{2}}} 4 - x'^2 -\sqrt{\dfrac{15}{4} - x'^2} dx'$

Let $x' = \dfrac{\sqrt{15}}{2}\sin(\theta) \implies dx' = \dfrac{\sqrt{15}}{2}\cos(\theta) d\theta \implies$

$\displaystyle\int \sqrt{\dfrac{15}{4} - x'^2} dx' = \dfrac{15}{8}\displaystyle\int 1 + \cos(2\theta) d\theta = \dfrac{15}{8}(\theta + \sin(\theta)\cos(\theta)) \implies$

$\displaystyle\int_{-\sqrt{\frac{7}{2}}}^{\sqrt{\frac{7}{2}}} \sqrt{\dfrac{15}{4} - x'^2} dx' =$ $\dfrac{15}{8}(\arcsin(\dfrac{2x'}{\sqrt{15}}) + \dfrac{2x'\sqrt{15 - 4x'^2}}{15})|_{-\sqrt{\frac{7}{2}}}^{\sqrt{\frac{7}{2}}}$

$= \dfrac{15}{4}\arcsin(\sqrt{\dfrac{14}{15}}) + \dfrac{\sqrt{14}}{4} \implies$ $A = \dfrac{31}{6}\sqrt{\dfrac{7}{2}} - \dfrac{15}{4}\arcsin(\sqrt{\dfrac{14}{15}})$

Using symmetry $\implies A_{T} = 2A = \dfrac{31}{3}\sqrt{\dfrac{7}{2}} - \dfrac{15}{2}\arcsin(\sqrt{\dfrac{14}{15}}) =$

$\dfrac{a}{b}\sqrt{\dfrac{c}{d}} - \dfrac{be}{d}\arcsin(\sqrt{\dfrac{cd}{be}}) \implies a + b + c + d + e = \boxed{48}$ .