Parabolas And Circles.

$D = r^2 = x^2 + (x^2 + 4 - y_{0})^2 \implies \dfrac{dD}{dx} = 2x(2x^2 + 9 - 2y_{0}) = 0$

$\implies x = 0, x = \pm\sqrt{\dfrac{2y_{0} - 9}{2}}$ and $x \neq 0 \implies x = \pm\sqrt{\dfrac{2y_{0} - 9}{2}} \implies$

$y_{0} = \dfrac{21}{4} \implies x = \pm\dfrac{\sqrt{3}}{2} \implies y = \dfrac{19}{4}$ .

Circle: $y_{2}(x) = -\sqrt{1 - x^2} + \dfrac{21}{4}$

Parabola: $y_{1}(x) = x^2 + 4$

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

Let $x = \sin(\theta) \implies dx = \cos(\theta) d\theta \implies$

$A_{R} = 2(-\dfrac{1}{2}\displaystyle\int_{0}^{\dfrac{\pi}{3}} (1 + \cos(2\theta)) d\theta +\dfrac{\sqrt{3}}{2})$

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

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