Lemoine-Lozada circles I made some rookie attempt to define the 4th Lemoine circle recently. The alternative name for this circle was suggested yesterday. Further investigation revealed a family of circles associated with the Brocard axis. Supposedly the name of Ehrmann-Lozada circles (or Lemoine-Lozada circles) can be safely attached to them.
Specifically, it appears that for each Lozada circle it is possible to construct at least one circle of the Lemoine type, with its center on the line that is going through the Symmedian point and the circumcenter of the triangle.
step 1: Let A'B'C' be the midheight triangle of ABC and let BA and CA be the orthogonal projections of B' and C' on BC, respectively. Build AB, CB, AC and BC similarly. These last six points lie on a circle with center X(9729). The circle is called the 1st Lozada circle. (César Lozada, March 17, 2016)
step 2: Circumcircles of the triangles BcAcX(6), AbCbX(6), BaCaX(6) cut the sides of the original triangle at six points,  that define the 1st Ehrmann-Lozada circle. Its center O_1 is not in the ETC.

Ehrmann-Lozada circles can be constructed in the similar manner for each(?) of the ten remaining Lozada circles:

*

*https://www.geogebra.org/geometry/pg9mrmzx      2 new circles of the Lemoine type

*https://geogebra.org/geometry/uqnpqjeh          2 new circles of the Lemoine type

*https://www.geogebra.org/geometry/t4jpn9dp    only 1 new circle of the Lemoine type

...
11. https://www.geogebra.org/m/kppjdcny      gives us only 1 new circle of the Lemoine type...
More links will be added soon. None of these points O_i is included in Clark Kimberling's encyclopaedia.
Finding a strict proof of existence of these new circles is the main question here. Bit I am also interested in what makes Lozada points/circles so special? (Taylor point belongs to the Brocards Axis as well, but for some reason it is impossible to construct an Ehrmann-Taylor circle in the same way!)  Are these Ehrmann-Lozada circles also Tucker circles?
 A: Once you have worked through even a handful of examples of a construction, it can be worthwhile to abstract things a bit in hopes of glimpsing a general principle.
A cyclic sextuple of points —two on each side-line of a triangle— is determined by a triple of of those points (one on each side-line). We can ask when such a triple gives rise to a secondary cyclic sextuple by the OP's construction, recapped below. (In what follows, "$X_Y$" and "$X_Y'$" indicate points on the side-line opposite triangle vertex $X$.)


*

*Given $A_B$, $B_C$, $C_A$, define $A_C$, $B_A$, $C_B$ as the "other" points where $\bigcirc A_BB_CC_A$ meets the (side-lines of) the triangle.

*Further, given a specified point $S$, let $B_C'$, $C_B'$ be the "other" points where $\bigcirc SB_AC_A$ meets the triangle; likewise, we have $C_A'$ and $A_C'$ via $\bigcirc SC_BA_B$, and $A_B'$ and $B_A'$ via $\bigcirc SA_CB_C$.


Definition. Triple $A_B$, $B_C$, $C_A$ is magical with respect to $S$ if the six derived points $A_B'$, $A_C'$, $B_C'$, $B_A'$, $C_A'$, $C_B'$ lie on a circle (which I'll call the triple's (First) Magic Circle).

(Note that the definition says nothing about the centers of the circumcircles involved. I'll get to that.)
For general $S$, the condition of magicality is complicated to express, involving a quartic relation. The condition happens to be replete with side-squares $a^2$, $b^2$, $c^2$, and the relation factors nicely when $S$ is the symmedian point, with barycentric coordinates $(a^2:b^2:c^2)$. For this case, we can state

Theorem. Triple $A_B$, $B_C$, $C_A$ is magical with respect to symmedian point $S$ if and only if
$$\frac{\alpha}{c^2}=\frac{\beta}{a^2}=\frac{\gamma}{b^2} \qquad\text{or}\qquad \frac{1-\alpha}{b^2}=\frac{1-\beta}{c^2}=\frac{1-\gamma}{a^2} \tag{1}$$ for the signed ratios
$$\alpha := \frac{|BA_B|}{|BC|}\qquad \beta :=\frac{|CB_C|}{|CA|} \qquad \gamma := \frac{|AC_A|}{|AB|}$$

Proof is by Mathematica-assisted symbol crunching (that I won't reproduce here), fueled mostly by power-of-a-point relations among all the points on the side-lines; the relations encoding the role of $S$ are what bog the calculations down. For the symmedian case, I suspect there's a not-unreasonable synthetic proof of $(1)$. (I should acknowledge: I don't doubt $(1)$, and much (most? all?) of what follows, already appears in the literature.)
If the first proportion in $(1)$ holds for $A_B$, $B_C$, $C_A$, then the second holds for $A_C$, $B_A$, $C_B$ (redefining $\alpha$, $\beta$, $\gamma$ appropriately). Consequently, the latter triple is also magical, which allows us to say

If triple $A_B$, $B_C$, $C_A$ is magical with respect to symmedian point $S$, then the six "other" points where $\bigcirc SA_BB_A$, $\bigcirc SB_CC_A$, $\bigcirc SC_AA_C$ meet the triangle are also concyclic, giving a Second Magic Circle.


Moreover, the construction is symmetric in the given points $A_B$, $B_C$, $C_A$ and derived points $A_B'$, $B_C'$, $C_A'$; thus, the derived triple is necessarily also magical (as is derived triple $A_C'$, $B_A'$, $C_B'$). We don't get any new circles, however; rather, we get the circumcircle of our original points (their "Zeroth Magic Circle"), and Second Magic Circle noted above. We effectively have a magical triple of circles: any one gives rise to the other two.
For a bit of metric specificity ... Define $\kappa_0$ as the (normalized) constant of proportionality in $(1)$.
$$\frac{\kappa_0}{a^2+b^2+c^2} \;:=\; \frac{\alpha}{c^2}=\frac{\beta}{a^2}=\frac{\gamma}{b^2}$$
Likewise defining $\kappa_1$ and $\kappa_2$ relative to the First and Second Magic Circles, we find
$$\kappa_{i+1} = \frac{3 - \kappa_i}{2 - \kappa_i}$$
Also, defining $k_i$ as the radius of $i$-th Magic Circle, we have
$$k_i^2 = \frac{r^2}{(a^2 + b^2 + c^2)^2}
  \left(\;(a^2 + b^2 + c^2)^2 (1-\kappa_i) \;+\; (a^2 b^2+b^2c^2+c^2a^2)\kappa_i^2\;\right)$$
where $r$ is the circumradius of $\triangle ABC$.
Recall that the definition of a magical triple of points made no assumptions about the centers of the original or derived circles; in particular, it seems to have ignored what seems to be the key property motivating OP's question. It happens that the symmedian case gives us that property for free:

If triple $A_B$, $B_C$, $C_A$ is magical with respect to symmedian point $S$, then its circumcenter lies on the Brocard axis (joining $S$ to the circumcenter $O$ of $\triangle ABC$).
Since the construction transfers magic to the derived points, we can say the Zeroth, First, and Second Magic Circles have centers $K_i$ on the Brocard axis. And, specifically,
$$\frac{|OK_i|}{|OS|}=\frac{\kappa_i}{2}$$

This, and the radius calculation, tell us how to construct a Magic Circle about any point on the Brocard axis. (Sanity check: $\kappa=0$ corresponds to $k=r$. The circle is the triangle's circumcircle. Points $A_B$, $B_C$, $C_A$ coincide with the vertices, so $\alpha=\beta=\gamma=0$.)
Also, one can confirm that each of the three Magic Circles is a Tucker circle. Converting the above results into Tucker/Brocard parameters is left as an exercise to the reader.

Okay, that's enough of that.

To address OP's particular interest in Lozada circles ...
I don't know if there's a unifying principle in that family of circles, or if "the $n$-th Lozada Circle" is simply "the $n$-th circle that Lozada happened to notice had a neat construction and a center on the Brocard axis". So, I can't say whether it's surprising —or maybe it's obvious— that members of the family are magical. Nevertheless, I'll explicitly confirm the magical nature of three cases illustrated by OP. (I'm afraid I don't have the patience to work through all eleven.)
To avoid a little clutter, define $\lambda := \kappa/(a^2+b^2+c^2)$ as the un-normalized proportionality factor in $(1)$. Revealing magic is simply a matter of showing that the value of $\lambda$ is symmetric in triangle metrics.

*

*First Lozada Circle: $A_B$, $B_C$, $C_A$ (and $A_C$, $B_A$, $C_B$) are orthogonal projections of mid-height points. Straightforward right-triangle trig yields
$$|BA_B|=r\sin^2C\sin A, \; |CB_C|=r\sin^2A\sin B, \; |AC_A|=r\sin^2B \sin C\;\to\; \lambda=\frac{1}{8r^2}$$

*Second Lozada Circle: $A_B$, $B_C$, $C_A$ (and $A_C$, $B_A$, $C_B$) are orthogonal projections of points where altitudes meet the circle having the orthocenter and centroid as a diameter. A coordinate bash gives
$$|BA_B| = \frac{8|\triangle ABC|^2}{3ab^2}, \quad |CB_C|=\cdots, \quad |AC_A|=\cdots \quad\to\quad \lambda = \frac{8|\triangle ABC|^2}{3a^2b^2c^2}$$

*Fourth Lozada Circle: $A_B$, $B_C$, $C_A$ (and $A_C$, $B_A$, $C_B$) are points where the side-lines meet lines through "opposite" sides of Pythagoras-esque squares erected on the triangle's sides. Here again, simple trig allows us to write
$$|BA_B|=-\frac{c}{\sin B}, \quad |CB_C|=\cdots, \quad |AC_A|=\cdots \quad\to\quad \lambda=-\frac{1}{2|\triangle ABC|}$$
(I don't know why OP asserts the existence of "only one" circle derived from the Fourth Lozada Circle. There are two.)

OP seems to be suggesting that the Taylor circle doesn't quite fit with this investigation. In fact, the Taylor Circle (rather, each of two triples of points it determines on the triangle) is magical:

*

*Taylor Circle: $A_B$, $B_C$, $C_A$ (and $A_C$, $B_A$, $C_B$) are orthogonal projections of the feet of triangle's altitudes onto the side-lines. Specifically, with $X_Y$ the projection of the foot from vertex $Y$ projected onto the side-line opposite vertex $X$, we find
$$|BA_B|= a\sin^2C \qquad |CB_C|=b\sin^2A \qquad |AC_A|=c\sin^2B \quad\to\quad \lambda=\frac{1}{4r^2}$$
Here's a figure showing a Taylor Circle (blue) and associated First and Second Magic Circles:



I'll stop typing now.
