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?
 A: As I promised, here is animation, that show locus point-by-point.


*

*Ellipses parameters: $a=2, b=1$. 

*Gray circle: circle with radius $3$.

*Point of contact "moves" along static ellipse (and along rolling one) with constant speed.


Locus has some deviation of perfect circle.

A: 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.
A: For the arc length, the middle expression should have a plus sign between the two terms.
In the last integral the limits should be root-5-over-3 to 2.
On the circle hypothesis, my crude experiment with cheese boxes wrapped in velcro, a = 69+or-1 mm, b = 43+or-1 mm, gave a diameter of 225 mm with a standard deviation of 5 mm.
The experiments form part of an article for a U.K. maths magazine targeted at secondary teachers (Mathematics in School). Since your posting I'd like to make teachers of  older students aware of Mathstackexchange. (Apart from the final integrals there is nothing in your proof which would not be in an A level further maths syllabus.) 
A: Just another style of counterexample.
We consider moment/instant, when ellipses are co-directed (image $2$) (see previous post).
1) Assume condition $(1)$ is true.
So, find point $M$, that divides arc $AB$ into 2 parts of equal length.
$$
\displaystyle \int\limits_0^{x_M} \sqrt{\dfrac{16-3x^2}{16-4x^2}}dx = \int\limits_{x_M}^2 \sqrt{\dfrac{16-3x^2}{16-4x^2}}dx  = \dfrac{1}{2} \int\limits_0^{2} \sqrt{\dfrac{16-3x^2}{16-4x^2}}dx.
$$
Coordinates of $M$: $x_M\approx 1.18894378$, because
$$
\displaystyle \int\limits_0^{x_M} \sqrt{\dfrac{16-3x^2}{16-4x^2}}dx \approx 1.211056028,
$$
$$
\displaystyle \int\limits_{x_M}^2 \sqrt{\dfrac{16-3x^2}{16-4x^2}}dx \approx 1.211056028.
$$
Then $y_M = \sqrt{1-x_M^2/4} \approx 0.80411639$.
2)
But condition $(2)$ is false now. $KL = 2KM = 2\sqrt{x_M^2+y_M^2} \approx  2.8706727$. It is far from $3$. Deviation is $4.31 \%$. 

When ellipses will have $a=5,b=1$, you'll find $KL \approx 5.5108044$, instead of $6$. Deviation  is $8.15 \%$ now.
