[Math] A circle tangent to an ellipse

conic sectionseuclidean-geometrygeometry

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.

Best Answer

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

Let us consider an ellipse $\mathcal{E}$ with equation $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$, i.e. $(x,y)=(a\cos\theta,b\sin\theta)$;
Let us consider a generic point $P\in\mathcal{E}$, i.e. some $\theta\in[0,2\pi)$;
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;
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 $x^2+y^2=(a+b)^2$;
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:

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 $\mathcal{E}$;
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:

Related Solutions

[Math] find the center of an ellipse given tangent point and angle

First, find the equation of the tangent line (using $\tan\alpha$). Then use orthogonal affinity in the coordinate system, along the $y$-axis: $$(x,y)\mapsto (x,\frac y3)$$ Then the ellipse goes to a circle, tangent goes to tangent, and you can use your formula. Then multiply back $y$ coordinate of the result centrepoint by $3$.

[Math] Plotting a skew ellipse

Part 1: \begin{align} x(t) &= \cos(\alpha) \cos(t) - h\sin(\alpha) \sin(t) \\ y(t) &= \sin(\alpha) \cos(t) + h\cos(\alpha) \sin(t) \end{align}

I don't understand the question for part 2. Are the two point supposed to be on the ellipse? If so, then $h$ and $\alpha$ cannot be determined, because there can multiple ellipses at the origin containing the same two points. (Quick proof: draw a (non-circular) ellipse $E$ at the origin. Rotate it by 30 degrees to get another ellipse $E'$. Now look at $E \cap E'$. It will typically contain four points. If any two of these are called $A$ and $B$, then both $E$ and $E'$ are possible solutions to the "find $h$ and $\alpha$" problem, but the $\alpha$ values for $E$ and $E'$ differ by 30 degrees.

Post-comment remarks

The squared distance from $(x(t), y(t))$ to $(0,0)$ is \begin{align} d(t)^2 &= \left[\cos(\alpha) \cos(t) - h\sin(\alpha) \sin(t)\right]^2 + \left[\sin(\alpha) \cos(t) + h\cos(\alpha) \sin(t)\right]^2 \\ &= \cos^2(\alpha) \cos^2(t) + h^2\sin^2(\alpha) \sin^2(t) - 2h\cos(\alpha) \cos(t)\sin(\alpha) \sin(t) + \sin^2(\alpha) \cos^2(t) + h^2\cos^2(\alpha) \sin^2(t) + 2h\cos(\alpha) \cos(t)\sin(\alpha) \sin(t) \\ &= (\cos^2(\alpha) + \sin^2(\alpha)) \cos^2(t) + h^2(\sin^2(\alpha) + \sin^2(t)) \sin^2(t) \\ &= \cos^2(t) + h^2 \sin^2(t) \end{align} This varies from $1$ (at $t = 0$) to $h^2$ (at $t = \frac{\pi}{2}$). Hence one (semi-) axis has length $1$ and the other has length $h$. If $h < 1$, then the major (semi-)axis will have length $1$.

Note that at $t = 0$, we have $$ \begin{align} x(0) &= \cos(\alpha) \cos(0) - h\sin(\alpha) \sin(0) &= \cos(\alpha) \ y(0) &= \sin(\alpha) \cos(0) + h\cos(\alpha) \sin(0) &= \sin(\alpha) \end{align} which is an angle $\alpha$ from the positive $x$-axis, as required.

It's possible that you still don't believe me, so here's some Matlab code:

function y = ellipseTest()
hList = [0.1, 0.3, 0.7];            % three different eccentricities
alphaList = [20, 50, 130] * pi/180; % three different major-axis angles
t = linspace(0, 2*pi, 100);         % sample values from 0 to 2pi
colorList = ['r', 'g', 'b'];
figure(gcf); 
clf;
hold on

for i = 1:3
   h = hList(i); alpha = alphaList(i); color = colorList(i);

    x = cos(alpha) * cos(t)  - h * sin(alpha) * sin(t);
    y = sin(alpha) * cos(t)  + h * cos(alpha) * sin(t); 
    plot(x, y, color)
end
hold off;
set(gca, 'DataAspectRatio', [1 1 1]);
figure(gcf);

and the resulting figure:

I find these pretty compelling.

Note that each pair of ellipses intersects in two points in the first quadrant, so if you picked one of these sets-of-two-points for your part 2 question, you'd find two different ellipses that fit them.

Best Answer

Related Solutions

[Math] find the center of an ellipse given tangent point and angle

[Math] Plotting a skew ellipse

Related Question