ellipse problem

The answer, for anyone interested,

Let $r$ be the radius of the circle. Then the center of the circle is $(0,b-r),$ so the equation of the circle is $x^2 + (y - (b - r))^2 = r^2.$
Solving for $x^2$ in $\frac{x^2}{a^2}+\frac{y^2}{b^2},$ we get $x^2 = \frac{a^2 (b^2 - y^2)}{b^2}.$
Substituting this into $x^2 + (y - (b - r))^2 = r^2,$ we get $\frac{a^2 (b^2 - y^2)}{b^2} + (y - (b - r))^2 = r^2.$
This simplifies to $(a^2 - b^2) y^2 + (2b^3 - 2b^2 r) y + (2b^3 r - a^2 b^2 - b^4) = 0.$
Since the ellipse and circle intersect at $(0,b),$ $y = b$ will be a root of the quadratic above. Thus, we can take out a factor of $y - b,$ which gives us $(y - b)((a^2 - b^2) y + a^2 b + b^3 - 2b^2 r) = 0.$
The solution to $(y - b)((a^2 - b^2) y + a^2 b + b^3 - 2b^2 r) = 0$ is $y_1 = \frac{2b^2 r - a^2 b - b^3}{a^2 - b^2}.$
If $y_1 < b,$ then the ellipse and circle also intersect at a point where the $y$-coordinate is $y_1$ (or the circle does not contain the ellipse). So, we need $y_1 = \frac{2b^2 r - a^2 b - b^3}{a^2 - b^2} \ge b.$
Isolating $r,$ we find $r \ge \frac{a^2}{b}.$
Therefore, the smallest possible radius is $\boxed{\frac{a^2}{b}}.$