I've got an idea for a simpler approach, but I've forgotten too much math to be able to actually try it.

The idea is to consider the set A of all circles that intersect the unit circle.

If you pick 3 random points inside the unit circle the probability that circle c ∈ A is the circle determined by those points should be proportional the length of the intersection of c's circumference with the unit circle.

The constant of proportionality should be such that the integral over all the circles is 1.

Then consider the set of all circles that are contained entirely in the unit circle. Integrate their circumferences times the aforementioned constant over all of these contained circles.

The ratio of these two integrals should I think be the desired probability.

I like this reasoning. Define a probability distribution on all circles of (x,y,r>0) based on how likely a given circle is. Then we can just sum the good circles and all the circles.

And the probability distribution is simple: a given (x,y,r) is as likely as its circumference in the unit circle.

Reasoning: Let C:(x,y,r) a given circle. We want to know how likely is it that the circle on 3 random points are close to it, closer than a given value d. (A d wide ball or cube around C in (x,y,r) space. Different shapes lead to diffferent constants but same for every circle.) The set of good 3 points is more or less the same as the set of 3 points from the point set C(d): make C's circumference d thick, and pick the 3 points from this set. Now not any 3 points will suffice, but we can hope that the error goes to 0 as d goes to 0 and there is no systematic error.

Then we just have to integrate.

ChatGPT got me the result 2/3, so it's incorrect. I guess the circumference must not be the right distribution.