Can an ellipse inscribed in a circle have a constant ratio of chords?

Question

This is a problem I've been trying to solve through simple algebra but the calculations get way too complicated so I thought I'd ask for some help.

Assume an ellipse inscribed in a circle centered at $(0, 0)$. Then, assume a line $\varepsilon$ passing through both at an angle $\phi$. Is the ratio of the chords $\frac{(AD)}{(BC)}$ constant for all angles $\phi$? Or perhaps just for the specific angles $\phi$ where $\varepsilon$ intersects both the circle and ellipse?

I actually solved this problem for the simpler case of the line $\varepsilon$ being parallel to the $x$-axis, ie the case where $y = \text{constant}$, like in the image:

It was pretty fun, and the result was that the ratio of the chords was equal to the ratio of the semi minor over the semi major axis (of course, in these problems the semi major axis and the circle's radius are equal). I tried to generalize the problem a bit by assuming any line $\varepsilon$, but got completely bogged down on the math, because I only know how to tackle such problems through algebra. The result I got seems to indicate that in the more general case, the ratio is not in fact constant. But I still wanted to ask in case I'm missing a more obvious and less calculations intensive method to approach this.

Answer 1

Consider the case where the the line passes through the origin. Then the chord $AD$ is a diameter of the circle, and so has the same length for any angle. The chord ABD can have any length between the minor and major axes of the ellipse, so the ratio is not constant.

Answer 2

Try an extreme case, such as a line $\varepsilon$ that passes through $(0,a)$ and $(a,0)$. Then $A=(a,0)$, and $C=D=(0,a)$. It's relatively straightforward to determine the coordinates of $B$, and from there you can compute that the chord ratio $AD/BC$ is quite different from $a/b$.

(I get a chord ratio of $\frac{a^2+b^2}{2b^2}$ in this extreme case, but my algebra might be wrong.)