Envelope of trilinear polars of isogonal conjugates of points on a circle

Given a circle, the envelope of the trilinear polars of the isogonal conjugates of points on the circle is a conic .

The conic, to be a parabola, needs that the given circle goes through the symmedian point $K$ .

Given a circle through $K$ centered at $Q$, call $F'$ the second intersection of $KQ$ and Jerabek hyperbola, and $F''$ the reflection of $F'$ in the midpoint of $K$ and $X_{5505}$. Then the focus $F$ of the parabola lies on line $KF''$ .

Then line joining $F'$ and $X_{5486}$ is parallel to the axis of the parabola .

The point $X_{5505}$ is the Kirikami concurrent circles image of $K$.

In general, let $P$ be a point in the plane of triangle $ABC$ . Let $H_a$ be the orthocenter of triangle $PBC$, and define $H_b$ and $H_c$ cyclically . Let $O_a$ be the circle through the points $A$, $H_b$, $H_c$, and define $O_b$ and $O_c$ cyclically . The circles $O_a$, $O_b$, $O_c$ concur at the Kirikami concurrent circles image of $P$.

The point $X_{5486}$ is the Kirikami Euler image of $K$ .

In general, let $P$ be a point in the plane of triangle $ABC$ . Let $H_a$ be the orthocenter of triangle $PBC$, and define $H_b$ and $H_c$ cyclically. The Euler lines of the triangles $AH_bHc$, $BH_cH_a$, $CH_aH_b$ concur at the Kirikami-Euler image of $P$.

Calculations with Mathematica (pdf version here)