A Circle inscribed in two intersecting parabolas!

Let $a > \dfrac{1}{\sqrt{2}}$ .

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

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

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

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

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

Using $y' = a^2 - x'^2 \implies D = d^2 = x'^2 + (a^2 - x'^2)^2 = x'^4 - (2a^2 -1)x'^2 + a^4 \implies$

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

$D = d^2 = \dfrac{4a^2 - 1}{4}$

and $a > \dfrac{1}{\sqrt{2}} \implies \dfrac{d^2D}{dx'^2} > 0 \implies$ the distance $d$ is minimized when $x' = \pm\sqrt{\dfrac{2a^2 - 1}{2}}$

$\implies A_{c} = \dfrac{2a^2 - 1}{4}\pi = \dfrac{a(3a + 1)}{4}\pi \implies a^2 - a - 1 = 0 \implies$ $a = \dfrac{1 \pm \sqrt{5}}{2}$ and dropping the negative root

$\implies a = \dfrac{1 + \sqrt{5}}{2} \approx \boxed{1.6180339887498948}$ .

Note: I could have let $|a| > \dfrac{1}{\sqrt{2}}$ which also works, but since $\dfrac{1 - \sqrt{5}}{2} \notin (-\infty, -\dfrac{1}{\sqrt{2}}) \cup (\dfrac{1}{\sqrt{2}}, \infty)$ there was no reason to extend the interval.