Question: > 
> 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?
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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.

Answer: 
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 $$\frac{3\sqrt{3}}{2}\pi$$
The problem has already been solved. If I put a *second* circle $x^2+(y+1)^2=1$ in the diagram, the problem is:

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> What is the ellipse of smallest area that can enclose two non-overlapping unit circles?
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On [Erich Friedman's Packing Centre](https://erich-friedman.github.io/packing/cirinel/) the following answer by James Buddenhagen is given:

[![Two circles in a minimum-area ellipse](https://i.stack.imgur.com/XLLYU.gif)](https://i.stack.imgur.com/XLLYU.gif)

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 $\frac{3\sqrt3\pi}2$.