Rolling ellipses I'm struggling to prove the following.
Set one ellipse in contact with a congruent one so that the minor axis of one is aligned with the major axis of the other. Now roll one round the other. The locus of the centre of the rolling ellipse is a circle centre the centre of the other, radius a + b.
Is there an obvious line of attack?

The locus is not a circle.
Counterexample.
Consider ellipses with semi-axes $a=2,b=1$. Let equation of static
ellipse is
$$
\dfrac{x^2}{2^2} +  \dfrac{y^2}{1^2} = 1.
$$
Here 3 steps are shown:

Suppose the locus is a circle (with radius  $r = a+b =3$).
Then must be an instant/moment (see $2$nd image), when ellipses are co-directed
(semimajor axes are parallel).
And 2 conditions must be true:
$$KM = LM = 3/2;\tag{1}$$
(since symmetry); and
$$len(BM) = len(AM),\tag{2}$$
where $len(...)$ is arc length. Yes, $len(BM) =^{\mbox{symmetry}} len(CM) =^{\mbox{rolling}} len(AM)$, since ellipse is rolling without slip. 

1).
Looking at condition $(1)$, let's find coordinates of point $M$.
$M$ belong to ellipse, $|KM|=3/2$.
$\left\{
\begin{array}{l}
x^2+4y^2 = 4; \quad (M \mbox{belong to ellipse});\\
x^2+y^2 = 9/4;\quad (|KM| = 3/2);
\end{array}
\right.
\quad \implies x = \sqrt{5/3},\; y = \sqrt{7/12}$.

2). Let's estimate $len(BM)$ and $len(AM)$.
Equation of arc $AB$ is $y = f(x) =\sqrt{1-x^2/4}$.
Note, that
$f'(x) = \dfrac{-x/2}{2\sqrt{1-x^2/4}} =
\dfrac{-x}{\sqrt{16-4x^2}}$.
$\sqrt{1 +(f'(x))^2} = \sqrt{\dfrac{16-4x^2}{16-4x^2} + \dfrac{x^2}{16-4x^2}} = \sqrt{\dfrac{16-3x^2}{16-4x^2}}$.
So,
$\displaystyle len(BM) = \int\limits_0^{\sqrt{5/3}} \sqrt{1+(f'(x))^2} dx = 
\int\limits_0^{\sqrt{5/3}} \sqrt{\dfrac{16-3x^2}{16-4x^2}}dx \approx 1.32081$ (wolfram alpha).
$\displaystyle len(AM) = \int\limits_{\sqrt{5/3}}^{2} \sqrt{1+(f'(x))^2} dx = 
\int\limits_{\sqrt{5/3}}^2 \sqrt{\dfrac{16-3x^2}{16-4x^2}}dx \approx 1.10131$ (wolfram alpha).
When condition $(1)$ is true, condition $(2)$ isn't true.
So, the locus isn't circle.
