Ellipses and Circles.

For the ellipse:

(1) $(j,j): j^2a + j^2b + j^2c + jd + je = 0$

(2) $(0,-j): jc - e = 0 \implies \boxed{c = \dfrac{e}{j}}$

(3) $2j,j): 4j^2a + 2j^2b + j^2c + 2jd + je = 0$

(4) $(2j,-j): 4j^2a - 2j^2b + j^2c + 2jd - je = 0$

Subtracting (4) from (3) $\implies 2jb + e = 0 \implies \boxed{b = -\dfrac{e}{2j}}$

Replacing $c = \dfrac{e}{j}$ and $b = -\dfrac{e}{2j}$ into equations (1) and (3) $\implies$

$2ja + 2d = -3e$

$4ja + 2d = -e$

$\implies \boxed{d = -\dfrac{5}{2}e}$ and $\boxed{a = \dfrac{e}{j}}$

$\implies \dfrac{1}{j}x^2 - \dfrac{1}{2j}xy + \dfrac{1}{j}y^2 - \dfrac{5}{2}x + y = 0$

$\implies 2x^2 - xy + 2y^2 - 5jx + 2jy = 0$

Solving for $y$ we obtain:

$y = \dfrac{1}{4}(x - 2j \pm \dfrac{1}{\sqrt{15}}\sqrt{384j^2 - (15x - 18j)^2})$

$y = \dfrac{1}{4}(x - 2j + \dfrac{1}{\sqrt{15}}\sqrt{384j^2 - (15x - 18j)^2})$ for the portion of the ellipse above the line $y = \dfrac{x - 2j}{4}$ .

Setting $384j^{2} - (15x - 18j)^{2} = 0 \implies x = \dfrac{2}{15}(9 \pm 4\sqrt{6})j$ are the points of intersection of the ellipse and the line $y = \dfrac{x - 2j}{4}$ .

Letting $a = \dfrac{2}{15}(9 - 4\sqrt{6})j$ and $b = \dfrac{2}{15}(9 + 4\sqrt{6})j$ the area of the ellipse is $A_{e} = \dfrac{2}{4\sqrt{15}}\displaystyle\int_{a}^{b} \sqrt{384j^{2} - (15x - 18j)^{2}})$

Letting $15x - 18j = \sqrt{384}j\sin(\theta) \implies dx = \dfrac{\sqrt{384}}{15}j\cos(\theta) \implies$

$I(\theta) = \dfrac{384}{15}j^2\displaystyle\int \cos^2(\theta) d\theta = \dfrac{384}{30}j^2\displaystyle\int(1 + \cos(2\theta)) d\theta = \dfrac{192}{15}j^2(\theta + \sin(\theta)\cos(\theta))$

$\implies A_{e} = \dfrac{96}{15^{\frac{3}{2}}}j^2(\arcsin(\dfrac{15x - 18j}{\sqrt{384}}) + \dfrac{(15x - 18j)\sqrt{384j^2 - (15x - 18j)^2})}{384})|_{a}^{b} =$ $\boxed{\dfrac{96\pi}{15^{\frac{3}{2}}}j^2}$ .

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

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

$(0,-j): x_{0}^2 + j^2 + 2jy_{0} + y_{0}^2 = r^2 \implies j(j + 2y_{0}), \:\ j \neq 0 \implies y_{0} = -\dfrac{j}{2}$

$(-3(j + 1),0): 9(j + 1)^2 + 6(j+ 1)x_{0} + x_{0}^2 + y_{0}^2 = r^2 \implies (j + 1)(3(j + 1) + 2x_{0}), \:\ j \neq -1 \implies x_{0} = -\dfrac{3}{2}(j + 1)$

$\implies r^2 = \dfrac{10j^2 + 18j + 9}{4}$

$\implies \boxed{A_{c} = (\dfrac{10j^2 + 18j + 9}{4})\pi}$

$\implies A = A_{c} - A_{e} = (\dfrac{10j^2 + 18j + 9}{4} - \dfrac{96}{15^{\frac{3}{2}}}j^2)\pi =$ $(\dfrac{2(5 * 15^{\frac{3}{2}} - 192)j^2 + 18 * 15^{\frac{3}{2}} + 9 * 15^{\frac{3}{2}}}{4 * 15^{\frac{3}{2}}})\pi = (\dfrac{625 * 15^{\frac{3}{2}} - 18816}{4 * 15^{\frac{3}{2}}})\pi$

$\implies (150\sqrt{15} - 384)j^2 + 270\sqrt{15}j + 18816 - 9240\sqrt{15} = 0$

Using the quadratic formula and dropping the negative root $\implies j = \boxed{7}$ .