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.
This may be also a picture proof, the reader may want to ignore below everything but the pictures and come with her or his own proof. But for didactic purposes, since we have a NMO problem and want to collect all points, there is also the story around the pics.
I need a lemma first, some points from the posted question are involved, notations are kept. I will introduce some more points, showing how the picture was realized.
Given is a triangle $\Delta ABC$, we draw its circumcenter $\odot(ABC)=\odot(O)$ with center $O$, the diameter perpendicular on $AC$ is $RS$, with $R$ on the same side of $AC$ as $B$. Consider on $AC$ points $M_1$, $N_1$
on $AC$, reflected w.r.t. the side bisector $RS$ of $AC$, denote by $m$ the common length $m=AM_1=CN_1$. Circles centered in $A$, respectively $C$ with radius $m$ further intersect the segments $AB$ in $M$, and respectively $CB$ in $N$. Let $\Omega$ be the circumcenter of the cirumcircle $\odot(BMN)$ of $\Delta BMN$.
Then:
- $R$ is also on the circle $\odot(\Omega)=\odot(BMN)$,
- $RM=RN$.
Proof: Draw parallels and perpendiculars through the circumcenters $O$ and $\Omega$ to the sides $BA$ and $BC$ of $\Delta ABC$. We obtain as in the picture two orange triangles with same hypotenuse $O\Omega$, and one of the legs is $m/2$ in each, the distance from the mid points of $AB$ and $MB$ for one triangle, respectively mid points of $CB$ and $NB$ for the other one.
These triangles are thus congruent, so $\Omega O$ is an angle bisector for an angle delimited by sides parallel to $BA$ and $BC$. So
$\Omega O$ is parallel to the angle bisector $BS$ of $\widehat{ABC}$. Since $\Omega O\|BS\perp BR$ we obtain that $\Omega O$ is also the side bisector of $BR$, thus $\Omega B=\Omega R$, so $R$ is also on the circle $\odot(\Omega)=\odot(BMN)$.
To see that $RM=RN$, compare $\Delta RNC$ and $\Delta RMA$. They have $RC=RA$, $NC=m=NA$, and $\widehat{NCR}=\widehat{BCR}=\widehat{BAR}=\widehat{MAR}$. So the two triangles are congruent. (And a rotation around $R$ with angle $\widehat{NRM}=\widehat{NBM}=\widehat{CBA}=\hat B$ brings one triangle into the other.)
$\square$
We are now in position to show the claimed result. Here there is a slightly changed way to introduce points, done in an equivalent manner. Since some other points, useful for the proof, are also introduced let us explicitly restate:
Problem: Let $\Delta ABC$ be given, consider $M,N,R,S$ as in the previous Lemma. Let $K=AC\cap MN$. Then $\bbox[yellow]{(1)}$ the quadrilaterals $AMRK$ and $CNRK$ are cyclic. Let $P^*$, $Q^*$ be the mid points of the arcs $\overset\frown{AM}$ of $\odot(AMRK)$, and respectively $\overset\frown{CN}$ of
$\odot(CNRK)$, that can be defined alternatively as the intersections of the angle bisector
$$\kappa$$
of $\widehat{AKM}=\widehat{CKN}$ with the two mentioned circles. Draw the interior and exterior angle bisectors of $\Delta ABC$ going either through $I$, its incenter, or through $I_B$, its $B$-excenter. Consider the intersections of these bisectors with $\kappa$:
$$
\begin{aligned}
P'&=\kappa\cap IA \ , & Q'&=\kappa\cap IC\ ,\\
P &=\kappa\cap I_BA\ , & Q &=\kappa\cap I_BC\ .
\end{aligned}
$$
Then $\bbox[yellow]{(2)}$
$\odot(APMP')$ is a circle centered in $P^*$, and
$\odot(CQNQ')$ is a circle centered in $Q^*$.
In particular we have $\bbox[yellow]{(3)}$:
$$
\begin{aligned}
RA&=RC\ ,\\
RM&=RN\ ,\\
RP^*&=RQ^*\ ,\\
RP&=RQ\ ,\\
RP'&=RQ'\ .
\end{aligned}
$$
Proof: For $\bbox[yellow]{(1)}$ - using the above Lemma:
$$
\widehat{RNK} =
\widehat{RNM} =
\widehat{RBM} =
\widehat{RBA} =
\widehat{RCA} \ .
$$
So $CNRK$ cyclic. Similarly $AMRK$ cyclic.
$\bbox[yellow]{(2)}$ take first from the picture only the piece involving the points
$K,C,N;Q,Q',Q*$ and the angle bisectors in $C,K$. Denote by $x,y$ half of the angle $\hat C$, respectively $\hat K$ in $\Delta CKN$. Then
$$
\begin{aligned}
\widehat{Q^*CQ'}
&=
\widehat{Q^*CN}
+
\widehat{NCQ'}
=
\widehat{Q^*KN}
+
\widehat{NCQ'}
=x+y=\widehat{Q'CK}
+
\widehat{Q'KC}
\\
&=\widehat{Q^*Q'C}\ .
\end{aligned}
$$
So $\Delta Q^*Q'C$ isosceles, $Q^*Q'=Q^*C$. The angle bisectors $CQ$, $CQ'$ are perpendicular, so $\Delta QCQ'$ has a right angle in $C$. Its mid point of the hypotenuse is the only point of this side at same distance from $C$ and $Q'$, so it is $Q^*$. We obtain thus $Q^*Q=Q^*C=Q^*Q'$. Since $Q^*$ is the mid point of the arc $\overset\frown{CN}$ of $\odot(CNRK)$, we can add also $Q^*N$ to the chain of equalities. So $Q^*$ is the center of the circle $\odot(CQNQ')$. The same happens also for $\odot(APMP')$.
We can attack $\bbox[yellow]{(3)}$. The obvious $R$-rotation moves $\Delta RNC\to\Delta RMA$, so take in the movement also corresponding objects, circles $\odot(RNC)\to\odot(RMA)$, arcs of them $\overset\frown{NC}\to\overset\frown{MA}$, and their mid points $Q^*\to P^*$. So
$$
RQ^*=RP^*\ ,
$$
so the perpendicular bisector of $P^*Q^*$ goes through $R$, and w.r.t. this bisector we have reflected pairs of points $(P^*,Q^*)$, and $(P,Q)$, and $(P',Q')$, the last two pairs since we know the distances on the line $\kappa=KPP^*P'QQ^*Q$.
$\square$
Best Answer
After some symbol-bashing in Mathematica, here are barycentric coordinates of key points:
$$\begin{align} A' &= (0 : a + b - c : a - b + c) \\[0.75em] A'' := \tfrac12(B'+C') &= \left(a +\frac{(a-b+c)(a+b-c)}{-a+b+c} : b : c\right) \\[0.75em] A''' &= (2 a + b + c : b : c ) \\[0.75em] O_A &= (2 a^3 : a^3 - b^3 + c^3 + a^2 b - a^2 c - 3 a b^2 - a c^2 - b^2 c + b c^2 - 4 a b c \\ &\phantom{(2 a^3\;}\quad : a^3 + b^3 - c^3 - a^2 b + a^2 c - a b^2 - 3 a c^2 + b^2 c - b c^2 - 4 a b c) \end{align}$$ The coordinates of the points where, say, $\bigcirc O_A$ (the one passing through incenter $X_1$) meets the side-lines of the circle are a little messy, so I'll omit them. Nevertheless, the six prescribed points are indeed cyclic, and the equation of their common circle has this barycentric form (in $u:v:w$ coordinates): $$\begin{align} 0 &= u^2 b c (-a + b + c) \cos A + v^2 c a (a - b + c) \cos B + w^2 a b (a + b - c) \cos C \\[0.75em] &\quad+ v w (-a^3 + b^3 + c^3 - 2 a^2 b - 2 a^2 c - b^2 c - b c^2) \\[0.75em] &\quad+ w u (-b^3 + c^3 + a^3 - 2 b^2 c - 2 b^2 a - c^2 a - c a^2) \\[0.75em] &\quad+ u v (-c^3 + a^3 + b^3 - 2 c^2 a - 2 c^2 b - a^2 b - a b^2) \end{align}$$
Converting to trilinear form (in $\alpha:\beta:\gamma$ coordinates), the equation can be manipulated into this: $$(\lambda \alpha + \mu\beta+\nu\gamma)(a\alpha+b\beta+c\gamma)+\kappa(a\beta\gamma+b\gamma\alpha+c\alpha\beta)=0$$ where $$\begin{align} \lambda &:= (-a + b + c)\cos A \\ \mu &:= (\phantom{-}a-b+c)\cos B\\ \nu &:= (\phantom{-}a+b-c)\cos C\\ \kappa &:= -2(a+b+c) \end{align}$$
As $\lambda:\mu:\nu$ are the trilinear coordinates for Kimberling center $X(219)$ ("$X(8)$-Ceva Conjugate of $X(55)$"), the circle is the Central Circle of that point. This answers the "alternatively" aspect of OP's question.
I don't know what Kimberling centers may lie on the circle. I don't even know how one would go about searching for them. Someone with more fluency in triangle-center lore may have some insights.
Incidentally, for the $\bigcirc O_A$ passing through $B''$, $B'''$, $C''$, $C'''$, etc, the corresponding six-point circle has trilinear form $$(\lambda'\alpha+\mu'\beta+\nu'\gamma)(a\alpha+b\beta+c\gamma)+\kappa'(a\beta\gamma+b\gamma\alpha+c\alpha\beta)=0$$ where $$\begin{align} \lambda' &:= a (-a + b + c)^2 \\ \mu' &:= b (\phantom{-}a - b + c)^2 \\ \nu' &:= c (\phantom{-}a + b - c)^2 \\ \kappa' &:= -8 a b c \end{align}$$ This circle is therefore the Central Circle of Kimberling's $X(220)$ ("$X(9)$-Ceva conjugate of $X(55)$").
I have not investigated whether there is a general connection between OP's construction, Ceva-conjugates, and/or $X(55)$.