Circle Problem

Let the center of the circle be the origin $O(0,0)$ and the line $BE$ is $b$ above $O$ . Then in coordinates, $E(a+2, b)$ and $C(-2, b-(a+2))$ . By Pythagorean theorem, we have:

$\begin{cases} OE^2: & (a+2)^2 + b^2 = r^2 \\ OC^2: & (-2)^2 + (b-(a+2))^2 = r^2 \end{cases}$

$\begin{aligned} \implies (a+2)^2 + b^2 & = 4 + (b-(a+2))^2 \\ & = 4 + b^2 - 2b(a+2) + (a+2)^2 \\ \implies 2b(a+2) & = 4 \\ b & = \frac 2{a+2} \end{aligned}$

Putting in $OE^2$ :

$\begin{aligned} (a+2)^2 + \frac 4{(a+2)^2} = r^2 \\ (a+2)^4 - 20(a+2)^2 + 4 & = 0 \\ \implies (a+2)^2 & = \frac {20+\sqrt{20^2-16}}2 = 10 + 4\sqrt 6 \\ a & = \boxed{\sqrt{10+4\sqrt 6} - 2} & \small \blue{\text{In response to Peter Csorba's comments}} \\ & = 2 + \sqrt 6 - 2 = \sqrt 6 \end{aligned}$

Therefore $\alpha + \beta + \gamma = 2+5+3 = \boxed{10}$ .

Let $(x_{0},y_{0})$ be the center of the circle.

$x_{0} - (a + 4))^3 + y_{0}^2 = r^2$

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

Subtracting the two equations we obtain:

$4(a + 2)x_{0} = 8(a + 2) \implies x_{0} = 2$ .

For $y_{0}:$

Using $x_{0} = 2$ we have:

$(a + 2)^2 + y_{0}^2 = r^2$

$4 + (y_{0} + (a + 2))^2 = r^2$

Subtracting the two equations we obtain:

$-2(a + 2)y_{0} = 4 \implies y_{0} = \dfrac{-2}{a + 2}$

$\implies$

$(a + 2)^2 + \dfrac{4}{(a + 2)^2} = 20 \implies (a + 2)^4 - 20(a + 2)^2 - 4 = 0 \implies$

$a = \pm\sqrt{10 \pm 4\sqrt{6}} - 2$

The only positive real root is $a = \sqrt{10 + 4\sqrt{6}} - 2 = \sqrt{2 * 5 + 2^2\sqrt{2 * 3}} - 2 =$

$\sqrt{\alpha * \beta + \alpha^{\alpha}\sqrt{\alpha * \lambda}}- \alpha \implies \alpha + \beta + \lambda = \boxed{10}$ .