Again Not the Deathly Hallows Symbol

If the radius of the circle is $r$ and the semimajor and semiminor axes of the ellipse are $a,b$ , then $r^2 = ab$ and the ellipse

$\frac{x^2}{a^2} + \frac{(y-b)^2}{b^2} \; = \; 1$ must be tangent to the two lines $y = 3r +2b \pm x\sqrt{3}$ , and hence $\begin{aligned} 3a^2 + b^2 & = \; (3r + b)^2 \\ 3a^2 & = \; 9r^2 + 6rb \\ a^2 & = \; 3ab + 2b\sqrt{ab} \end{aligned}$ so that $(1 - 3u)^2 = 4u^3$ where $u = \tfrac{b}{a}$ . Thus we deduce that $u=\tfrac14$ (the other solution $u=1$ may be a solution of this last equation, but the corresponding $b=a$ is not a solution of the previous equation - squaring an equation can create bogus solutions). Thus the eccentricity is $e = \sqrt{1-u^2} = \tfrac14\sqrt{15}$ , which makes the answer $\boxed{968245}$ .