Look at Euler line or Euler circle, and these are just examples. There are numerous properties in the triangle, many involving the orthocenter. And there are litterally hundreds of special points.
Some even say it's a sin to spend too much time looking for such properties. :-)
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 $X(55)$.
Best Answer
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.
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
(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:
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.