Ending Ellipses

If you place the ellipse so that its center is at the origin, its equation is $\cfrac{x^2}{24^2} + \cfrac{y^2}{6^2} = 1$ , which rearranges to $x^2 + 16y^2 = 576$ .

The large circle with a radius of $R$ will have a center of $(0, 6 - R)$ , so it has an equation of $x^2 + (y - 6 + R)^2 = R^2$ .

Substituting $x^2 = 576 - 16y^2$ , the equation becomes $576 - 16y^2 + (y - 6 + R)^2 = R^2$ and rearranges to $(y - 6)(y - \frac{2}{15}(R - 21)) = 0$ , which means that the ellipse and the circle intersect at $y = 6$ and $y = \frac{2}{15}(R - 21)$ .

For the ellipse to be inscribed and not intersect the circle, $\frac{2}{15}(R - 21) = y \geq 6$ , which solves to $R \geq 96$ .

The small circle with a radius of $r$ will have a center of $(24 - r, 0)$ , so it has an equation of $(x - 24 + r)^2 + y^2 = r^2$ .

Substituting $y^2 = \frac{1}{16}(576 - x^2)$ , the equation becomes $(x - 24 + r)^2 + \frac{1}{16}(576 - x^2) = r^2$ and rearranges to $(x - 24)(y - \frac{8}{15}(51 - 4r)) = 0$ , which means that the ellipse and the circle intersect at $x = 24$ and $x = \frac{8}{15}(51 - 4r)$ .

For the ellipse to be inscribed and not intersect the circle, $\frac{8}{15}(51 - 4r) = x \geq 24$ , which solves to $r \leq \cfrac{3}{2}$ .

Therefore, the ratio of the minimum possible large circle radius to maximum possible small circle radius is $\cfrac{96}{\frac{3}{2}} = \cfrac{64}{1}$ , so $a = 64$ , $b = 1$ , and $a + b = \boxed{65}$ .