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.
 A: Parameterise the ellipse as $x=a\cos t$, $y=b\sin t$.
This gives $x'=-a\sin t$, $x''=-a\cos t$, and $y'=b\cos t$, $y''=-b\sin t$.
Using this formula from Wikipedia, the curvature is then
$$
\begin{align}
\kappa &= \frac{x'y''-y'x''}{\left({x'}^2+{y'}^2\right)^{3/2}}\\
&=
{
ab
\over
(a^2\sin^2t+b^2\cos^2t)^{3/2}
}
\end{align}
$$
and at the top of the ellipse we have $t=\pi/2$ so this becomes $ab/a^3=b/a^2$. The radius of curvature is $1/\kappa=a^2/b$, so the circle has radius $a^2/b$.
A: First solution, without calculus.
Let the circle have its centre at $(0,-k)$ and touch the ellipse at $(0,b)$, so that its equation is
$$
x^2+(y+k)^2=(b+k)^2.
$$
Substitute here $x^2=a^2-{a^2\over b^2}y^2$ from the ellipse equation and expand to get:
$$
{a^2-b^2\over b^2}y^2-2ky+b^2-a^2+2bk=0.
$$
This always has $y=b$ as a solution, while the other solution is:
$$
y={b^3-a^2b+2b^2k\over a^2-b^2}.
$$
For $k>0$ small enough this will give the ordinate of two other intersection points, but increasing the value of $k$ we eventually get $y>b$, which must be discarded because it entails $x^2<0$. In that case there are no other intersections between circle and ellipse and the ellipse entirely lies inside the circle. The limiting value of $k$ is then obtained for $y=b$: plugging that into the above equation and solving for $k$ we thus find the limiting value
$$
k={a^2-b^2\over b}
$$
and the radius of the smallest circle containing the whole ellipse:
$$
r=b+k={a^2\over b}.
$$

Second solution, purely geometric.

The radius of curvature at a point $P$ of an ellipse is the harmonic
mean of the diameters of the circles tangent to the ellipse at $P$ and
passing through either focus of the ellipse

See here for a proof.
In our case the foci are at the same distance $a$ from $P=(0,b)$ and the circles described above are coincident. Their diameter $PQ$ is the hypothenuse of a right triangle (see figure) and by similar triangles we then have $PQ:a=a:b$, whence $PQ=a^2/b$. This is then the requested radius of curvature.

A: For a rundown of this answer including diagrams, please check out this Desmos graph.
Let us change the framing of this question. Consider instead a circle centred at the origin at radius $r$. Then if we translate the ellipse $r$ units upwards, its centre will be at $(0, r)$, with the vertical semi-major axis having length $b$. It thus follows that we can parametrise the ellipse in these new coordinates as:
$$(x(t), y(t)) = (a \cos t, b \sin t + r - b) \tag{$0 ≤ t < 2 \pi$}$$
Consider the squared distance function to the origin $r(t) = (a \cos t)^2 + (b \sin t + r - b)^2$. We know that $r(\pi/2)$ must equal $r^2$ exactly so that the $x$-coordinate is $0$ for the tangent point, after discounting $t = 3 \pi/2$: this will be our sanity check. We also know that this distance must be locally constant in the neigbourhood of $t = \pi/2$, so moreover, $f'(\pi/2) = 0$ (a local maximum in fact). Looking at the diagram, with $t$ going anticlockwise around the ellipse, $r(t)$ is increasing at first, then stabilises at $t = \pi/2$ before it decreases, which makes sense.
However, the information that $f'(\pi/2) = 0$ is not sufficient to express $r$ in terms of $a, b$. Without finding $f'(t)$ explicitly, we note that there will be a factor of $\cos t$ in both $(a \cos t)^2$, and $\cos t$ will appear too in the second term due to the chain rule $(\sin t)' = \cos t$. However, $\cos(\pi/2) = 0$ so we end up with $0 = 0$, a tautology.
The natural step would be then to observe $f''(t)$ around $t = \pi/2$. And the second derivative is enough information for the ellipse to have the same radius of curvature as its tangent circle.
Because a circle's distance to the origin is constant. Not only is its first derivative $0$, but its second derivative must equal $0$, and so must its third derivative. And any function wishing to approximate a circle around a point with a given radius of curvature must satisfy at least the first and second conditions (the higher derivatives do specify additional information which the radius of curvature cannot capture alone).
Now let us see if using the second derivative: $r''(\pi/2) = 0$ yields any new information. And it does indeed:
$$r'(t) = a^2 (2 \cos t)(-\sin t) + 2(b \sin t + r - b)(b \cos t)$$
$$= -a^2 \sin 2t + b^2 \sin 2t + 2b(r-b)\cos t$$
$$r''(t) = -2a^2 \cos 2t + 2b^2 \cos 2t+ 2b(b - r) \sin t$$
hence $r''(\pi/2) = 0$ yields $-2a^2 \cdot -1 + 2b^2 \cdot -1 + 2b(b-r) = 0$, or:
$$2a^2 - 2b^2 + 2b^2 - 2br = 0 \implies 2br = 2a^2 \implies \boxed{r = \frac{a^2}{b}}.$$
A: 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 $r = \frac{a^2}{b}$
