<s>[INST] How to find the radius of the smallest circle such that the inner ellipse is tangent to the circle at exactly one point? Consider the ellipse
$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1,$ where $a > b > 0.$ As a function of $a$ and $b,$ find the radius of the smallest circle that contains the ellipse, is centered on the $y$-axis, and which intersects the ellipse only at $(0,b).$

I have no clue how to solve this. The only thing I've thought of is that maybe it somehow involves the fact that for any $x$, the corresponding $y$ value of the circle is bigger than the corresponding $y$ value of the ellipse. But I haven't made much progress with that.
 [/INST] A solution without calculus.
Eliminating $x$ in the set $\cases{\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\\ x^2+(y-y_0)^2= r^2}$ we obtain
$$
\left(1-\frac{a^2}{b^2}\right)y^2-2y_0y+y_0^2-r^2=0
$$
now solving for $y$ to determine the intersections we have
$$
y = \frac{-y_0\pm\sqrt{y_0^2-(\frac{a^2}{b^2}-1)(r^2-y_0^2-a^2)}}{(\frac{a^2}{b^2}-1)}
$$
but we need tangency so this implies on an unique solution so the condition is
$$
y_0^2-(\frac{a^2}{b^2}-1)(r^2-y_0^2-a^2)=0
$$
solving for $r$ we get
$$
r = \frac{\sqrt{a^4-a^2b^2+a^2y_0^2}}{\sqrt{a^2-b^2}}
$$
but $r=b-y_0$ and thus we obtain $y_0 = \frac{b^2-a^2}{b}$ and $r = \frac{a^2}{b}$
</s>