Let $F$ be the focus of a conic, and let $P_1 Q_1$ and $P_2 Q_2$ be two focal chords of the conic, passing through $F$. Circles are drawn with $P_1 Q_1$ and $P_2 Q_2$ as diameters. Prove that the radical axis of these circles passes through a fixed point.
Note: This is a generalization of the result in this link: Prove that the common chord passes through the origin.
Here's a proof for the case of an ellipse. An analogous reasoning should work for a hyperbola. (See the link given in the question for the case of a parabola).
The supposed fixed point $M$ must be, by symmetry, the intersection of all radical axes with the line $FF'$ through the foci of the ellipse. We can consider, for instance, the circles having as diameter the major axis $AB$ and the latus rectum $DE$ of the ellipse (see figure below) with their radical axis $LK$ intersecting $FF'$ at $M$. From: $$ LM^2=FL^2-FM^2=CL^2-(CF+FM)^2 $$ one finds: $$ FM=a(1-e^2){e\over2}, $$ where $a$ is the length $CA$ of the semi-major axis, $e=CF/CA$ is the eccentricity of the ellipse and we used the result $FD=a(1-e^2)$ for the semi-latus rectum of an ellipse.
We want to show that point $M$ has the same power with respect to any circle having a focal chord through $F$ as diameter, thus proving that it lies on all radical axes.
Let's take then a generic focal chord $PQ$, with $FP\le FQ$ (see figure below). If $\theta$ is the acute angle formed by $FP$ with the major axis, then by the cosine law (applied to triangles $FPF'$ and $FQF'$) we have: $$ FP={a(1-e^2)\over1+e\cos\theta}, \quad FQ={a(1-e^2)\over1-e\cos\theta}. $$ If $O$ is the midpoint of $PQ$, the power of point $M$ with respect to the circle of diameter $PQ$ is $OM^2-OP^2$. We have: $$ OP={FP+FQ\over2}={a(1-e^2)\over1-e^2\cos^2\theta} $$ and $$ OF=OP-FP=a(1-e^2){e\cos\theta\over1-e^2\cos^2\theta}. $$ We can then compute $OM$ with the cosine law: $$ OM^2=OF^2+FM^2+2\,OF\cdot FM\cos\theta= \\ =a^2(1-e^2)^2\left( {e^2\cos^2\theta\over(1-e^2\cos^2\theta)^2}+ {e^2\over4}+ {e^2\cos^2\theta\over1-e^2\cos^2\theta} \right) $$ and finally obtain: $$ OM^2-OP^2=a^2(1-e^2)^2\left({e^2\over4}-1\right). $$ As the power of $M$ doesn't depend on $\theta$, then $M$ lies on the radical axis of any pair of circles having a focal chord as diameter.
