Inscribed Circles!

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

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

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

$\implies P:(\dfrac{\sqrt{3}}{2},0)$ and $P':(-\dfrac{\sqrt{3}}{2},0)$ .

Let $y = ax^2 + bx + c \implies c = \dfrac{9}{4}$

and

$-\dfrac{3}{2} = \dfrac{3}{4}a + \dfrac{\sqrt{3}}{2}b$

$-\dfrac{3}{2} = \dfrac{3}{4}a - \dfrac{\sqrt{3}}{2}b$

$\implies a = -2$ and $b = 0 \implies y = -2x^2 + \dfrac{9}{4}$

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

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

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

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

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