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.
Best Answer
There is a survey article by Berndt, Choi, and Kang devoted to the set of 58 Ramanujan's problems. They indicate that the questions had originally appeared in the problems section of the Journal and apparently the editors published readers' solutions in subsequent issues.
Concerning your question 1, let me just quote from the Introduction to the survey:
An elementary solution to the specific geometric problem you've mentioned can be found in Ramanujan's Notebooks, Part III by Berndt (Springer, 1991, pp. 244-246). The problem stems from Ramanujan's work on modular equations of degree 3...