circle on ellipse

In this problem the ellipse and circle are fixed. The ellipse has center $E$ on the origin, its semi-minor axis $r$ is on the $y$-axis, and its semi-major axis $R$ is on the $x$-axis. The circle has center $O$ and radius $c$. The circle and the ellipse intersect at point $P$.

Given $E, O, R, r$, and $c$, is it possible to determine $\angle z=\angle EOP$?

When $P$ lies on the $x$ or $y$-axis, the angle is $0$. But when it lies anywhere else, as in the picture, I'm stumped as to how to easily calculate $z$.

Thank you.


The problem posed is algebraically intractable.

If instead of O you know $P$, then you can find the angle after extending line OP to meet the major axis at $F$.

Label the intersection on the right of the ellipse with its major axis $G$. Let
$\angle GFO$ be $\alpha$ and $\angle GEO$ be $\beta$; then if we can determine $\alpha$ and $\beta$, $\angle z = \alpha - \beta$.

But if the projections of $EP$ along the major and minor axes are $(x,y)$ respectively, then $$ \tan \alpha = \frac{yR^2}{xr^2} $$ and $$ \tan \beta = \frac{y+c\sin\alpha}{x+c \cos \alpha} $$ Take the respective arctangents and subtract.

For the problem posed, you have $O$ so $\beta$ is trivially $\tan^{-1}\frac{y_0}{x_0}$, but to get $\alpha$ you would need to solve for $x, y, \alpha$ in the equations $$ \left\{ \begin{array}{c} \tan \alpha = \frac{yR^2}{xr^2} \\ x + c \cos \alpha = x_0 \\ y + c \sin \alpha = y_0 \end{array} \right. $$ You can use trig identities to eliminate $\alpha$ and solve the last equation for $y$ in terms of $x$, but the remaining equation involves square roots of two expressions, one of which has $(x-x_0)^2$ in a denominator, and when you group and square twice you end out with an 8-th degree equation which is much too tough to solve.

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Given the data, can you not determine P as the point of intersection of the circle and the ellipse? Then, you can use the method by @MarkFischler.

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  • $\begingroup$ Unlike Nehorai, I do see how this is a hint for the OP. However, this hint is very weak. Can you strengthen it? $\endgroup$ – Rory Daulton Jan 2 '16 at 15:54
  • $\begingroup$ Well, a statement was made that this problem was not tractable. I believe otherwise. Expanding on my comment above, a circle and an an ellipse will intersect at up to four points (which can be determined, presumably, by looking at the quadratic equations that represent each figure and solving for the points of intersection. There are many on the net that discuss the mathematics (e.g.: math.stackexchange.com/questions/655619/…). In this case, the points are coincidental -- the point P. Once you know P, and given E and O, you can compute the angle Z. $\endgroup$ – Dr. Parasolian Jan 3 '16 at 17:38

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