A 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 $x' = \pm\sqrt{\dfrac{7}{2}}$

$\implies A_{c} = \dfrac{15}{4}\pi = \dfrac{a}{b}\pi \implies a + b = \boxed{19}$ .