Inscribed Circle!

Using the rotation equations:

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

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

For $x^2 + 2\sqrt{3}xy + 3y^2 + 2\sqrt{3}x - 2y = 0$ we have:

$x'\cos^2(\theta) - x'y'\sin(2\theta) + y'\sin^2(\theta) + 3(x'^2\sin^2(\theta) + x'y'\sin(2\theta) + y'\sin^2(\theta)) +$

$2\sqrt{3}(x'y'\cos(2\theta) + (x'^2 - y'^2)\sin(\theta)\cos(\theta)) + 2\sqrt{3}(x'\cos(\theta) - y'\sin(\theta))$

$- 2(x'\sin(\theta) + y'\cos(\theta)) = 0$

Setting $x'y'$ term to zero we obtain:

$2\sin(2\theta) + 2\sqrt{3}\cos(2\theta) = 0 \implies \tan(2\theta) = -\sqrt{3} \implies 2\theta = \dfrac{2\pi}{3} \implies \theta = \dfrac{\pi}{3}$

Replacing $\theta = \dfrac{\pi}{3}$ in the above equation $\implies 4x'^2 - 4y' = 0 \implies y' = x'^2$

Similarly for $-2x^2 - 4\sqrt{3}xy - 6y^2 + 2\sqrt{3}x - 2y + 9 = 0$ we obtain $y' = -2x'^2 + \dfrac{9}{4}$ .

Working from $x'y'$ system and minimizing the distance $OP$ we have:

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

$x' = \pm\sqrt{\dfrac{2y'_{0} - 1}{2}} \implies \dfrac{2y'_{0} - 1}{2} + (\dfrac{2y'_{0} - 1}{2} - y'_{0})^2 = 1$

$\implies \dfrac{2y'_{0} - 1}{2} = \dfrac{3}{4} \implies y'_{0} = \dfrac{5}{4} \implies A:(0,\dfrac{9}{4})$ and $x' = \pm\dfrac{\sqrt{3}}{2}$

$\implies P:(\dfrac{\sqrt{3}}{2},0)$ and $Q:(-\dfrac{\sqrt{3}}{2},0)$ and the equation of the circle in the $x'y'$ system is

$x'^2 + (y' - \dfrac{5}{4})^2 = 1$

For $R_{1}$ and $R_{2}:$

$I = 2\displaystyle\int_{0}^{\dfrac{\sqrt{3}}{2}} (-2x'^2 + \dfrac{9}{4}) - (\dfrac{5}{4} - \sqrt{1 - x'^2}) \:\ dx' =$

$2\displaystyle\int_{0}^{\dfrac{\sqrt{3}}{2}} (-2x'^2 + 1 + \sqrt{1 - x'^2}) \:\ dx'$ .

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

$I = 2((-\dfrac{2x'^3}{3} + x')|_{0}^{\dfrac{\sqrt{3}}{2}} + (\dfrac{1}{2}(\theta + \dfrac{1}{2}\sin(2\theta)))|_{0}^{\dfrac{\pi}{3}}) =$

$2(\dfrac{3\sqrt{3}}{8} + \dfrac{\pi}{6}) = \dfrac{3\sqrt{3}}{4} + \dfrac{\pi}{3}$

$\implies \boxed{A_{1} = \pi - I = \dfrac{2\pi}{3} - \dfrac{3\sqrt{3}}{4}}$

For $R_{3}:$

For $A_{2} = 2\displaystyle\int_{0}^{\dfrac{\sqrt{3}}{2}} ((\dfrac{5}{4} - \sqrt{1 - x'^2}) - x'^2) \:\ dx =$

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

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

$A_{2} = 2(\dfrac{5}{4}x' - \dfrac{x'^3}{3}|_{0}^{\dfrac{\sqrt{3}}{2}} - \dfrac{1}{2}(\theta + \dfrac{1}{2}\sin(2\theta))|_{0}^{\dfrac{\pi}{3}}) =$

$2(4\dfrac{\sqrt{3}}{8} - \dfrac{\sqrt{3}}{8} - \dfrac{\pi}{6}) = \dfrac{3\sqrt{3}}{4} - \dfrac{\pi}{3}$

$\boxed{A_{2} = \dfrac{3\sqrt{3}}{4} - \dfrac{\pi}{3}}$

$\implies$ the desired area $A = A_{1} + A_{2} = \dfrac{\pi}{3} \approx \boxed{1.04719755}$