Angle arising from circle rolling along an 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.
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.