Find the smallest ellipse containing a family of ellipses of the same shape Given the usual definition of ellipse
$\mathcal{E}(A,c):=\{ x\in \mathbb{R}^2 | (x-c)^\top A (x-c) \le 1 \}$ with $A=A^\top>0 $,
I am trying to solve the following problem
Problem: Given an ellipse $\mathcal{E}_2:=\mathcal{E}(A/\alpha^2,c_2)$ with $\alpha>1$ find the smallest value of $\alpha$ such that $\forall\, c_1$, s.t., $\| c_1 - c_2 \| \le d$, $\mathcal{E}_1:=\mathcal{E}(A,c_1) \subseteq \mathcal{E_2}$.
In other words, I trying to find what is the minimum dilatation factor such that $ \mathcal{E_2}$ contains any possibile $\mathcal{E_1} $ (remind that dividing the matrix $A$ by $\alpha^2$, the semiaxes are enlarged of a factor $\alpha$).
My first attemp was to consider this problem geometrically. Consider a circle centered in $c_2$ with radius $d$. All the possible ellipses $\mathcal{E}_1$ have their centre inside this circle. In the attempt to find the worst case, I consider the four ellipses whose center is on the circumpherence and one of the two axis is tangent to the circumpherence (see Fig.1).

Let's call the semi-axes of a generic ellipse $\mathcal{S(E)}$.
By Fig. 1, it is clear that the containing ellipse $\mathcal{E}_2$ must satisfy the following conditions

*

*$\min{\mathcal{S}(\mathcal{E}_2)} \ge d+\min{\mathcal{S}(\mathcal{E}_1)} $, satisfied by $\alpha \ge \dfrac{(d+\min{\mathcal{S}(\mathcal{E}_1)})^2}{\min{\mathcal{S}(\mathcal{E}_1)}^2} $

*$\max{\mathcal{S}(\mathcal{E}_2)} \ge d+\max{\mathcal{S}(\mathcal{E}_1)} $ satisfied by $\alpha \ge \dfrac{(d+\max{\mathcal{S}(\mathcal{E}_1)})^2}{\max{\mathcal{S}(\mathcal{E}_1)}^2} $
If $ \alpha \ge \dfrac{ (d+\min{\mathcal{S}(\mathcal{E}_1)})^2 }{ \min{\mathcal{S} (\mathcal{E}_1)}^2} \ge \dfrac{(d+\max{\mathcal{S}(\mathcal{E}_1))^2} }{ \max{\mathcal{S}(\mathcal{E}_1)}^2 }$ both condition are satisfied.
Since I am looking for the smallest containing ellipse $\mathcal{E}_2$, i.e., the smallest $\alpha$, I choose $\alpha = \dfrac{ (d+\min{\mathcal{S}(\mathcal{E}_1)})^2 }{ \min{\mathcal{S} (\mathcal{E}_1)}^2}$ (see Fig.2)

Drawing all the possible ellipses $\mathcal{E}_1$, it seems that this value of $\alpha$ solves the problem (see Fig. 3)

Do you have any suggestion about how formally prove this conjecture?
Maybe the solution could be easily extended to the 3D case...
Thank you all for your attention!
 A: The value of $\alpha$ is determined by the condition of tangency between the two ellipse $E_1(c_1,1)$ and $E_2(c_2\alpha) $  At the tangency point $r$ we have
$(r - c_1)^T A (r - c_1) = 1 $
$(r - c_2) A (r - c_2) = \alpha^2 $
$A (r - c_1) = \kappa A (r - c_2) $
for some $\kappa \gt 0 $
The last equation implies
$ r - c_1 = \kappa (r - c_2) $
so that
$ (1 - \kappa) r = c_1 - \kappa c_2 $
which simplifies further to
$ r = \dfrac{ c_1 - \kappa c_2 }{1 - \kappa} $
Plug this expression for $r$ into the very first equation
$ ( c_1 (\kappa - 1) + c_1 - \kappa c_2 )^T A ( c_1 (\kappa - 1) + c_1 - \kappa c_2) = (1 - \kappa)^2 $
This reduces to
$ ( c_1 - c_2 )^T A (c_1 - c_2) = \bigg( 1 - \dfrac{1}{\kappa} \bigg)^2$
Whose solution is
$ | 1 - \dfrac{1}{\kappa} | = \sqrt{ (c_1 - c_2)^T A (c_1 - c_2) } $
Noting that $\kappa \lt 1 $ because the first ellipse is smaller than the second ellipse, then
$ \kappa = \dfrac{1}{ \sqrt{(c_1 - c_2)^T A (c_1 - c_2) } + 1 } $
Finally evaluate $r$ with the found value of $\kappa$ and substitute it into the second equation, and this will give $\alpha$.
The following shows an example that I worked out using the above equations.  The smaller fixed ellipse is the blue one centered at $c_1 = (3,6)$, while the scalable one is centered at $c_2 = (5,4) $ and shown in pink.  The orange ellipse is the smallest scaled version of the pink ellipse that covers the first ellipse.

