Areas and Rotations.

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 $\cos(2\theta) = 0 \implies 2\theta = \dfrac{\pi}{2} \implies \theta = \dfrac{\pi}{4}$

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

Using the equations of rotation with $\theta = \dfrac{\pi}{4}$ we obtain

$-\dfrac{a}{\sqrt{2}} = \dfrac{1}{\sqrt{2}} x'- \dfrac{1}{\sqrt{2}}y'$

$\dfrac{a}{\sqrt{2}} = \dfrac{1}{\sqrt{2}}x' + \dfrac{1}{\sqrt{2}}y'$

$\implies x' = 0, y' = a \implies$ $(-\dfrac{a}{\sqrt{2}},\dfrac{a}{\sqrt{2}}) \rightarrow (0,a)$ .

Using $y' = x'^2$ and center $O':(0,a)$ and $P':(x',y') = (x',x'^2) \implies$

$D = r^2 = x'^2 + (x'^2 - a)^2 \implies \dfrac{dD}{dx} = 2x'(2x'^2 + 1 - 2a) = 0$

$x' \neq 0 \implies x' = \pm\sqrt{\dfrac{2a - 1}{2}}$ and the radius $r = 1 \implies 4a - 1 = 4 \implies$

$a = \dfrac{5}{4} \implies x' = \dfrac{\sqrt{3}}{2} \implies y' = \dfrac{3}{4} \implies$ the equation of the circle is

$x'^2 + (y' - \dfrac{5}{4})^2 = 1$ and the portion of the circle needed is $\boxed{y' = \dfrac{5}{4} - \sqrt{1 - x'^2}}$ .

The area $A = 2\displaystyle\int_{0}^{\dfrac{\sqrt{3}}{2}} (\dfrac{5}{4} - \sqrt{1 - x'^2} - x'^2) dx$

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

$A = 2(-\dfrac{1}{2}\displaystyle\int_{0}^{\dfrac{\pi}{3}} (1 + \cos(2\lambda)) d\lambda + (\dfrac{5}{4}x' - \dfrac{x'^3}{3})|_{0}^{\frac{\sqrt{3}}{2}}) =$

$2(-\dfrac{1}{2}(\lambda + \dfrac{1}{2}\sin(2\lambda))|_{0}^{\frac{\pi}{3}} + \dfrac{\sqrt{3}}{2})$

$= \dfrac{3\sqrt{3}}{4} - \dfrac{\pi}{3} = \dfrac{a\sqrt{a}}{b} - \dfrac{\pi}{a} \implies a + b = \boxed{7}$ .

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Note: In the $xy$ system using the equations of rotation with $\theta = \dfrac{\pi}{4}$ the center is $(-\dfrac{5}{4\sqrt{2}},\dfrac{5}{4\sqrt{2}}) \implies$

the equation of the circle(in the xy system) is $(x + \dfrac{5}{4\sqrt{2}})^2 + (y - \dfrac{5}{4\sqrt{2}})^2 = 1$

and $P:(\dfrac{-2\sqrt{3} - 3}{4\sqrt{2}}, \dfrac{3 - 2\sqrt{3}}{4\sqrt{2}})$ and $Q:(\dfrac{2\sqrt{3} - 3}{4\sqrt{2}}, \dfrac{3 + 2\sqrt{3}}{4\sqrt{2}})$