Dusting off your geometry book ...
Given any two points on a circle the center lies on to the perpendicular bisector of the chord between them. Here the perpendicular bisector passes through the midpoint $(1/2)((0,0)+(4,2))=(2,1)$. The slope of the chord is clearly $+1/2$ so, by the "negative reciprocal rule" the perpendicular line has slope $-2$. So the center lies on
$y-1=-2(x-2), y=-2x+5$
Since the center also lies on $y=1-x$ we then have for the center:
$1-x=-2x+5, x=4, y =-3$
meaning the center is $(4,-3)$. The radius should now be easy to figure out given that $(4,2)$ is on the circle centered at $(4,-3)$, and the equation of the circle follows.
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.
Best Answer
Let me first restate the problem: what is the maximum size of a concyclic subset of $\{1,2,3,4,5\}^2 \subseteq \mathbb{R}^2$?
Let's first consider some easy bounds on the problem. Since there is a circle through any three noncollinear points, the answer is at least 3. For an upper bound, notice that the point set is contained in five lines, and a line intersects a circle in at most two points. So the answer is at most 10.
The example you found is natural to come up with, since the point set has 8-fold symmetry. There are other similar examples, like a circle of radius $\sqrt{5}$ centered at the center of the point set. So the answer is at least 8.
This question may seem difficult because there are infinitely many circles, but since the point set is finite, there are actually a tractable number of circles that contain many points. I claim that 10 is not the answer. Suppose there was a circle going through ten points -- it must go through exactly two in each row and two in each column. The number of choices for these points in the first row is $\binom{5}{2} = 10$, and given these points, the number of choices for another point in the same row as the lower point is $4$. So you only need to check 40 circles. One could come up with longer arguments excluding more circles, but 40 is not too big.
In fact, you can use the exact same argument to show that 9 is not the answer. So 8 is the correct answer.