What values gives the minimum area of the ellipse? 
If the ellipse $\dfrac{x^2}{A}+\dfrac{y^2}{B}=1$ is to enclose the circle $x^2+y^2=2y$, what values of $A,B>0$ minimize the area of the ellipse?

So far I've put the circle equation into the standard form: $x^2+(y-1)^2=1$, and I know that $A,B$ represents the length of semi-major and semi-minor axis. I'm not sure how to make sure the circle is enclosed in the ellipse.
 A: We require the two conics to touch each other. Eliminating $x^2$ leads to the quadratic equation $$(A-B)y^2+2By-AB=0$$
Assuming $A\neq B$, applying the condition that the discriminant is zero leads to the equation $$A^2-AB+B=0$$
We now need to minimize the area $\Delta=\pi\sqrt{AB}$.
Therefore we can differentiate $$\Delta^2=\pi^2AB=\pi^2\frac{A^3}{A-1}$$
Setting the derivative to zero will give $$A=\frac 32, B=\frac 92$$
It is readily seen that this will provide the minimum area since there is no maximum. Therefore the semiaxes are $$\sqrt{A}=\sqrt{\frac 32}, \sqrt{B}=\frac{3}{\sqrt{2}}$$
The minimum ellipse area is then $$\frac{3\sqrt{3}}{2}\pi$$
A: The problem has already been solved. If I put a second circle $x^2+(y+1)^2=1$ in the diagram, the problem is:

What is the ellipse of smallest area that can enclose two non-overlapping unit circles?

On Erich Friedman's Packing Centre the following answer by James Buddenhagen is given:

The ellipse that solves both the problem above and the original question has semi-major axis $\frac{3}{\sqrt2}$, semi-minor axis $\sqrt\frac32$ and area $\frac{3\sqrt3\pi}2$.
A: From the equation
$x^2+(y-1)^2=1$,
the circle has center
$(0, 1)$
and radius $1$.
Therefore the points on it
with minimum and maximum
$x$ and $y$ values are
$(0, 0), (1, 1),
(0, 2), (-1, 1)$.
Since the ellipse
has center at the origin,
it has to have a
max $y$ value of
at least $2$
and a max $x$ value
of at least $1$.
For the equation
$\dfrac{x^2}{A}+\dfrac{y^2}{B}=1$,
this means that
$A \ge 1$
and
$B \ge 4
= 2^2$.
However,
if the ellipse
has greater curvature
at the top
than the circle,
it will intersect the circle.
The radius of curvature 
of the ellipse
with $A=1, B=4$ is
$\frac{a}{b}
=\frac14
$,
which is smaller than
that of the circle,
which is $1$.
So,
we have to modify the
values of $A$ and $B$
so that the ellipse
is tangent to the circle.
An easy way to do this
is to make the ellipse
a circle of radius $2$,
so the values are
$A=B=4$.
Another possibility
is to make 
$A$ slightly larger
and
$B$ larger
so that the ellipse
is tangent to the circle
at two points.
I think that,
for any $A> 1$,
the value of $B$ 
that makes this happen
could be determined,
but I do not feel
like working this out.
The desired answer
would be the one
that minimizes
$AB$.
