New method for constructing six-point circles? 
*

*I would like to present a circle associated with a triangle that can
be constructed as follows:
Let A' be the first intersection point of the bisector of
angle A with the inscribed circle of the triangle ABC and
define B' and C' cyclically. I is the incenter of
the triangle ABC. Then circumcircles of the triangles A'B'I,
B'C'I,  A'C'I  intersect the sides of the original triangle at six points, that are always concyclic:
Geogebra dynamic sketch
I wonder what triangle centres (if any) might be lying on this "red circle" and what  is known about it?
Point Z (the center of the "red circle" that is passing through six points) is not included into the Clark Kimberling's Encyclopedia...   Mathworld.wolfram.com  doesn't mention it as well.  
P.S.
The same construction can be also applied to other
remarkable circles:
Geogebra dynamic sketch for the Taylor point/circle. 
Geogebra dynamic sketch for a Cevian circle of the Symmedian point. <slightly different construction principle was applied here, but overall it is the same idea. X(6) is not the center of the Symmedian circle>
Geogebra dynamic sketch for the nine point circle
So perhaps this principle holds true for all central circles and there might be    some sort of a generalisation?
 A: Here's an outline of a proof of a generalization that covers all of OP's six-point variants as well as giving significance to situations where the circumcircles don't actually meet the sides of the triangle.

Consider $\triangle ABC$, an arbitrary point $O'$, and points $A'$, $B'$, $C'$ on $\overleftrightarrow{O'A}$, $\overleftrightarrow{O'B}$, $\overleftrightarrow{O'C}$. There exists a circle $\kappa$ whose radical axes with $\bigcirc O'B'C'$, $\bigcirc O'C'A'$, $\bigcirc O'A'B'$ are the side-lines $\overleftrightarrow{BC}$, $\overleftrightarrow{CA}$, $\overleftrightarrow{AB}$, respectively.


In other words, $\kappa$, $\bigcirc O'B'C'$, and $\overleftrightarrow{BC}$ belong to an Apollonian family; likewise for the other two circle-line pairs. Since the radical axis of intersecting circles contains their common chord (aka, the line through their points of intersection), circle $\kappa$ will contain all circumcircle-line intersections that happen to appear. In the context of OP's stated result, this says that the points of intersection are concyclic.
I'm certain there's an elegant synthetic proof exploiting Apollonian families and such, but I got the result via rampant coordinate-bashing in Mathematica.
I set $O'$ at the origin, established coordinates for $A$, $B$, $C$, $A'$, $B'$, $C'$, and took $\kappa$ to be a circle with center $K:=(h,k)$ and radius $r$.
A convenient thing about the radical axis of two circles is that its equation can be obtained by subtracting the (monic) equations of the circles themselves. So, we have this system (where the "$\text{constant}$"s account for the fact that each line equation is determined only up to a scalar multiple).
$$\begin{align}
(\text{monic eqn of $\bigcirc O'B'C'$})-(\text{monic eqn of $\kappa$})=\text{constant}_1\cdot(\text{eqn of $\overleftrightarrow{BC}$}) \\
(\text{monic eqn of $\bigcirc O'C'A'$})-(\text{monic eqn of $\kappa$})=\text{constant}_2\cdot(\text{eqn of $\overleftrightarrow{CA}$}) \\
(\text{monic eqn of $\bigcirc O'A'B'$})-(\text{monic eqn of $\kappa$})=\text{constant}_3\cdot(\text{eqn of $\overleftrightarrow{AB}$}) \\
\end{align} \tag{1}$$
Equating coefficients of $x$ and $y$ and the constant terms gives a system of nine equations for unknowns $h$, $k$, and $r$. This seems hopelessly overdetermined, but the collinearities $\overline{O'A'A}$, $\overline{O'B'B}$, $\overline{O'C'C}$ work appropriate magic that allows us ultimately to solve, simplify like mad, and find

$$\begin{align}
K \;&= (bb'-cc')A^\perp + (cc'-aa')B^\perp + (aa'-bb')C^\perp \\[1em]
16\,|\triangle ABC|^2\,r^2 \;&= \phantom{+16\,}|BC|^2 (aa'-cc') (aa'-bb') \\[4pt]
&\phantom{=}+ \phantom{16\,}|CA|^2 (bb' - aa') (bb' - cc')\\[4pt]
&\phantom{=}+ \phantom{16\,}|AB|^2 (cc' - bb') (cc' - aa')\\[4pt]
&\phantom{=}+ 16\, |\triangle ABC| \, \left(\,aa' |\triangle O'BC| + bb' |\triangle O'CA| + cc' |\triangle O'AB|\,\right)
\end{align}$$
where $a:=|O'A|$, $a':=|O'A'|$, etc, and $(x,y)^\perp:=(-y,x)$.

(Provided $r^2$ is non-negative, which I'll leave for the reader to explore,) These give us our target circle $\kappa$, as desired. $\square$

It's worth noting that there's an even-more-general version of this result that increases the Apollonian factor:

Consider points $O$, $A$, $B$, $C$, $O'$, $A'$, $B'$, $C'$ such that $\square OAA'O'$, $\square OBB'O'$, $\square OCC'O'$ are cyclic. There exists a circle $\kappa$ such that $\kappa$, circumcircle $\bigcirc OBC$, and circumcircle $\bigcirc O'B'C'$ belong to an Apollonian family (ie, they have a common radical axis), and likewise the other circumcircle pairs.

The earlier result corresponds to taking $O$ as the "point at infinity", so that circumcircle $\bigcirc OBC$ is simply the line $\overleftrightarrow{BC}$ (itself the radical axis of the corresponding Apollonian family), etc.
Okay, full disclosure: I haven't actually crunched the symbols on this generalization. However, my GeoGebra sketch is pretty compelling; here's an instance where $\kappa$ is a six-point circle.

This is probably a standard lemma in the study of Apollonian families; perhaps even an "obvious" one. However, I've been staring at coordinate soup for so long that I'm not in the proper frame of mind to think synthetically. I'll leave that, too, to the reader.
