Circles and Ellipses.

Using $ax^2 + bxy + cy^2 + dx + ey = 0$ and the points I chose above to generate the ellipse.

(1) $(0,-4): 4c - e = 0 \implies \boxed{c = \dfrac{e}{4}}$

(2) $(8,4): 16a + 8b + 4c + 2d + e = 0$

(3) $(8,-4): 16a - 8b + 4c + 2d - e = 0$

Subtracting (3) from (2) we obtain: $8b + e = 0 \implies \boxed{b = -\dfrac{e}{8}}$

(4) $(5,8): 25a + 40b + 64c + 5d + 8c = 0$

Replacing $c = \dfrac{e}{4}$ and $b = -\dfrac{e}{8}$ into (2) and (4) $\implies$

$16a + 2d = -e$

$25a + 5d = -19e$

$\implies \boxed{a = \dfrac{11e}{10}}$ and $\boxed{d = -\dfrac{93e}{10}}$

$\implies \dfrac{11}{10}x^2 - \dfrac{1}{8}xy + \dfrac{1}{4}y^2 - \dfrac{93}{10}x + y = 0 \implies$ $44x^2 - 5xy + 10y^2 - 372x + 40y = 0 \implies 10y^2 + 5(x - 8)y + 44x^2 - 372x = 0$

Solving for $y$ we obtain:

$y = \dfrac{x - 8}{4} \pm \dfrac{1}{20}\sqrt{\dfrac{5}{347}}\sqrt{2207744 - (347x - 1448)^{2}}$ .

$y(x) = \dfrac{x - 8}{4} +\dfrac{1}{20}\sqrt{\dfrac{5}{347}}\sqrt{2207744 - (347x - 1448)^2}$ for the portion of the ellipse above the line $y = \dfrac{x - 8}{4}$ .

Setting $2207744 - (347x - 1448)^2 = 0 \implies x = \dfrac{1448 \pm 448\sqrt{11}}{347}$ are the points of intersection of the ellipse and the line $\dfrac{x - 4}{4}$ .

Letting $a = \dfrac{1448 - 448\sqrt{11}}{347}$ and $b =\dfrac{1448 + 448\sqrt{11}}{347}$ the area of the ellipse is $A_{e} = 2\displaystyle\int_{a}^{b} y(x) - \dfrac{x - 8}{4} \:\ dx =$
$\dfrac{1}{10}\sqrt{\dfrac{5}{347}}\displaystyle\int_{a}^{b} \sqrt{2207744 - (347x - 1448)^{2}} dx =$ $\dfrac{1}{2\sqrt{5}\sqrt{347}}\displaystyle\int_{a}^{b} \sqrt{2207744 - (347x - 1448)^2} dx$ .

For $I(x) = \displaystyle\int \sqrt{2207744 - (347x - 1448)^2} dx$

Let $347x - 1448 = \sqrt{2207744}\sin(\theta) \implies dx = \dfrac{\sqrt{2207744}}{347}\cos(\theta)$

$\implies I(\theta) = \dfrac{2207744}{347}\displaystyle\int \cos^2(\theta) d\theta = \dfrac{2207744}{694}\displaystyle\int (1 + \cos(2\theta)) d\theta = \dfrac{2207744}{694}(\theta +\sin(\theta)\cos(\theta))$

$\implies I(x) = \dfrac{2207744}{694}(\arcsin(\dfrac{347x - 1448}{\sqrt{2207744}}) + \dfrac{347x - 1448}{2207744}\sqrt{2207744 - (347x - 1448)^2}) \implies$

$A_{e} = \dfrac{551936}{(347)^{\frac{3}{2}}\sqrt{5}}(\arcsin(\dfrac{347x - 1448}{\sqrt{2207744}}) + \dfrac{347x - 1448}{2207744}\sqrt{2207744 - (347x - 1448)^2})|_{a}^{b} = \dfrac{1103872}{(347)^{\frac{3}{2}}\sqrt{5}}(\dfrac{\pi}{2}) =$

$\boxed{\dfrac{551936\pi}{(347)^{\frac{3}{2}}\sqrt{5}}}$

For the circle $(x - x_{0})^2 + (y - y_{0}) = r^2$ :

$(0,0): x_{0}^2 + y_{0}^2 = r^2$

$(0,-4): x_{0}^2 + 16 + 8y_{0} + y_{0}^2 = r^2 \implies y_{0} = -2$

$(-3,0) 9 + 6x_{0} + x_{0}^2 + y_{0}^2 = r^2 \implies x_{0} = -\dfrac{3}{2}$

$\implies r^2 = \dfrac{25}{4} \implies \boxed{A_{c} = \dfrac{25}{4}\pi}$

$\implies A = A_{e} - A_{c} = (\dfrac{551936}{(347)^{\frac{3}{2}}\sqrt{5}} - \dfrac{25}{4})\pi =$ $(\dfrac{(2^2)^{5} * 539}{(347)^{\frac{3}{2}}\sqrt{5}} - (\dfrac{5}{2})^2)\pi =$ $(\dfrac{(\alpha^{\alpha})^{\beta}\lambda}{\omega^{\frac{\gamma}{\alpha}}\sqrt{\beta}} - (\dfrac{\beta}{\alpha})^{\alpha})\pi$ $\implies \alpha + \beta + \lambda + \gamma + \omega = \boxed{896}$ .