# A circle tangent to an ellipse

A friend of mine showed me the following problem:

Let $\cal E$ be an ellipse whose semi major axis has length $a$ and semi minor axis has length $b$. Let $\ell_1, \ell_2$ be two parallel lines tangent to $\cal E$. Let $\cal C$ be the circle tangent to $\ell_1$, $\ell_2$, and $\cal E$. Prove that the distance between the centers of $\cal C$ and $\cal E$ is equal to $a+b$.

So far I managed to prove that if we draw the tangent line $k_1$ through $\cal E \cap \cal C$ and the tangent $k_2$ to $\cal E$ parallel to $k_1$ then the circle tangent to $k_1, k_2, \ell_1$ is tangent to $\cal E$ as well.

I'm stuck. I'd like to see some proofs, preferably non-analytic ones.

• Wild guess: consider the simplest possible case where the ellipse isn't "tilted". If that works, show that rotating the ellipse can't change the distance. – barrycarter Jul 15 '17 at 14:50
• I found this problem in "sangaku". – Takahiro Waki Jul 15 '17 at 14:53
• @TakahiroWaki Could you please provide a more precise reference? Thanks. – timon92 Jul 15 '17 at 15:32
• (+1) Very nice question. Sangaku problems are well-known to be incredibly deep sometimes. – Jack D'Aurizio Jul 15 '17 at 16:47
• @Narasimham: animation added to my answer. – Jack D'Aurizio Jul 16 '17 at 2:40

I have a roadmap for a simple solution through analytic geometry / trigonometry:

1. Let us consider an ellipse $\mathcal{E}$ with equation $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$, i.e. $(x,y)=(a\cos\theta,b\sin\theta)$;
2. Let us consider a generic point $P\in\mathcal{E}$, i.e. some $\theta\in[0,2\pi)$;
3. Let us compute the slope of the tangent $\tau_{P}$ through $P$ and consider that the parallel tangent is $\tau_{-P}$, since in a conic the midpoints of parallel chords are aligned along a line through the center of the conic;
4. Let $R_\theta$ be half the distance between $\tau_P$ and $\tau_{-P}$ and $\ell_\theta$ the line through the original parallel to $\tau_P$. Let $C_\theta$ be an intersection between $\ell_\theta$ and the circle $x^2+y^2=(a+b)^2$;
5. In order to check that $\mathcal{E}$ and the circle with radius $R_\theta$ centered at $C_\theta$ are tangent it is enough to check that a discriminant equals zero.

Here there are some ideas for a purely Euclidean solution:

1. Let $Q=\mathcal{C}\cap\mathcal{E}$. First Claim: the parallelogram having sides $\tau_P,\tau_{-P},\tau_Q,\tau_{-Q}$ has its vertices on a fixed ellipse $\mathcal{E}'$, having the same foci as $\mathcal{E}$;
2. Let $U,V$ the vertices of the previous parallelogram on $\tau_Q$. Second Claim: $C_\theta U$ and $C_\theta V$ are orthogonal tangents to $\mathcal{E}'$, so $C_\theta$ lies on the orthoptic of $\mathcal{E}'$ and has a constant distance from $O$.

Indeed, the first claim follows from Poncelet's porism and the second claim is just a matter of angle chasing.

Here it comes an animation, too:

• I think these two claims lead to a solution. You have proved that $OC_\theta$ does not depend on $\theta$, so it is enough to check what happens in one particular case - e.g. the case when $C_\theta$ lies on the major axis of $\cal E$, but in this case the problem is trivial. Am I missing something? – timon92 Jul 15 '17 at 17:25
• @timon92: you are not missing anything, you are completely right. – Jack D'Aurizio Jul 15 '17 at 17:37
• Thank you very much. This was incredible. – timon92 Jul 15 '17 at 17:42
• @hypergeometric: Geogebra allows that, too, one needs just a bit of patience. – Jack D'Aurizio Jul 16 '17 at 14:40
• @TakahiroWaki: no, it is not. I do not see how you can prove that $OC_\theta$ is constant with an affine transformation. If you are able to do that you are probably able to prove Poncelet's porism with affine transformations only: I do not believe that's possible. – Jack D'Aurizio Jul 19 '17 at 8:02