O(n) if you sweep. In the code I sent, I reordered the sweep to give better best-case performance.

Let’s denote the radius of the wanted circle R, the points we want to encapsulate will be “given points”, number of given points is N. Also let us assume that all the given points are distinct.

Find a set of points that have distance R from at least two given points. Then one of the optimal circles will have center in one of these points. There is O(N^2) of such points and evaluating each such point takes O(N). This gives us O(N^3) algorithm.
(When we find no such point, we can obviously encapsulate at most one point, and some center is trivial to find.)

Why it works?

If we have a circle encapsulating M points (M>1), there will be a circle encapsulating at least M points with center that lies R far from two points.

[Proof:
First for R-distance from one point:

If we have a circle encapsulating M points whose center doesn't lie R far from any point, we can keep moving it in any direction as long as no point falls out of it. And a point falls out of it only when it was R far from it and now we are moving it further.

For the second point it is similar -- we are R far from one point (P), and we can move the center point on the circle with center in P and diameter R until some other point slips from our encapsulating circle (to have such point, we need the M>1 condition). This again happens only when a point that was R-far from the center point gets a bit further.]

BTW the clique finding is NP-complete so in some cases it can be worse than the heatmap algorithm (unless it for some reason isn’t NPC for this kind of graphs, or 122 points is just too little for it to happen)