Consider a circle of radius $r$ and an ellipse of semiaxes $a>b$, both centered at the origin $O$, with area of the circle and the ellipse being equal, such as in the Figure. The ellipse has an angle $\hat{ROP}$ between its major semiaxis and coordinate axis $x$. Distance $OP = r$. Also $OR=r$, and $R$ is both on the ellipse and the circle
Given a horizontal line $m$ that passes through point $P$, we know that the acute angle between that line and the semiaxis $a$ is equal to $\hat{ROP}$. I can see graphically that line $m$ intersects the ellipse at point $Q$, which coincides with the endpoint of semiaxis $b$. However I need to prove this algebraically: Prove that the distance between the intersection of $m$ with the ellipse and the origin, is $b$.
Any suggestions are welcome.
By comparison of areas:
$$\pi r^2=\pi ab$$
Let us take the principal axes of the ellipse as coordinate axes: the coordinates $(x,y)$ of intersection point $R$ are solutions of the following system:
$$\begin{cases}x^2&+&y^2&=&ab& Circle\\\dfrac{x^2}{a^2}&+&\dfrac{y^2}{b^2}&=&1&Ellipse\end{cases} \iff \begin{cases}x^2&+&y^2&=&ab\\b^2 x^2&+&a^2y^2&=&a^2b^2\end{cases}\tag{1}$$
whose solutions are:
$$x^2=\dfrac{a^2b}{a+b} \ \ \text{then, by symmetry between} \ x,y, \ \ y^2=\dfrac{b^2a}{a+b}$$
Let us consider, among the four solutions, the coordinates of $M$ situated in the fourth quadrant: $x>0,y<0$, in order to "stick" to the case represented on the graphics, i.e.,
$$x=\sqrt{\dfrac{a^2b}{a+b}}, \ \ \ \ \ \ y=-\sqrt{\dfrac{b^2a}{a+b}}\tag{2}$$
OR and OP will be parallel if and only if slope(OR) = slope(QP).
Using (2)
$$slope(OR)=\dfrac{y}{x}=-\sqrt{\dfrac{b}{a}},$$
$$slope(QP)=-\dfrac{OQ}{OP}=-\dfrac{b}{r}=-\dfrac{b}{\sqrt{ab}}$$
which are indeed identical.