# What is the probability that the circumscribed circle $B$ of three randomly selected points within a circle $A$ will be included in circle $A$?

If three points are randomly selected within a circle $$A$$ (area uniformly), what is the likelihood that the circumscribed circle $$B$$ of these points will be completely contained within circle $$A$$?

Using Monto Carlo Method, implement it in Mathematica, the result is roughly $$0.40$$.

But, Is there an analytic solution? how to get it?

n = 10^6;

Table[circle = Circumsphere[RandomPoint[Disk[], 3]];
(Norm[circle[[1]]] + circle[[2]]) <= 1, n]
//Count[#, True] &
//(#/n) &
// N // AbsoluteTiming

{71.4554, 0.400337}
{74.6617, 0.40019}


# Update

I found a post on another website, showing the result is:

$$P_n=\frac{2^{n^2} \pi^{\frac{n}{2}-1} n !\left(\frac{n \Gamma\left(\frac{n}{2}\right)}{\Gamma\left(\frac{n+1}{2}\right)}\right)^n \Gamma\left(\frac{n^2+1}{2}\right)^2}{\left(n^2+n\right) !}$$ where $$n$$ means $$n$$-dimension unit-sphere.

for $$n=2$$, it produces $$\frac{2}{5}$$

But I can't understand the derivation steps. Any help would be appreciated.

This is not a complete answer stating the numeric value. This is an outline of the steps needed along the way.

Your question didn't say so, but I assume you have your points independently and uniformly distributed inside the circle. In that case the size of the circle doesn't really matter, so you might as well assume the circle $$A$$ is the unit disk.

So you can use e.g. https://mathworld.wolfram.com/DiskPointPicking.html or https://en.wikipedia.org/wiki/N-sphere#Generating_random_points to describe the probability distribution for picking two points $$P$$ and $$Q$$ on the disk, in a way where you can integrate over all possible pairs of points and weigh each combination with its corresponding density. You might also simplify your life somehow by assuming that you pick $$P$$ but then rotate the coordinate system in such a way that $$P$$ lies on the positive $$x$$ axis, without loss of generality. You might for example get something like

$$P=\frac1{2\pi}\int_0^1\int_0^1\int_{-\pi}^\pi f\left(\begin{pmatrix}\sqrt t\\0\end{pmatrix}, \begin{pmatrix}\sqrt u\cos\varphi\\\sqrt u\sin\varphi\end{pmatrix}\right) \mathrm d\varphi\,\mathrm du\,\mathrm dt$$

where $$f(P,Q)$$ is the probability that a circle through $$P$$ and $$Q$$ will be contained within the unit disk.

The limit between a circumscribed circle lying within the unit circle and a circumscribed circle crossing the unit circle is a circumscribed circle which touches the unit circle in a single point. This is a special case of Appolonius' problem, and https://en.wikipedia.org/wiki/Special_cases_of_Apollonius%27_problem#Type_5:_One_circle,_two_points has instructions for how to construct such touching circles. I also wrote an answer in the past where I used Lie geometry to compute specific solutions to Apollonius' problem.

For your given points $$P$$ and $$Q$$ you would in the general case get two circles through these points and touching the unit circle. A circle though $$P$$, $$Q$$ and a third point $$R$$ lies within the unit circle if and only if $$R$$ lies within one of the Apollonius circles but outside the other. That's the yellow area in the following figure:

The probability of the point $$R$$ falling inside the yellow area is equal to the ratio between that yellow area and the total area of the unit disk. So at this stage you'd compute the two touching circles, Sum their area but make sure to exclude the area where they overlap, and that's your function $$f$$ for your probability integral.

Given that the area computations are likely to make heavy use of transcendental functions, I assume that actually coming up with a closed formula for the whole integral is very unlikely. Instead you would end up doing numerical approximations of some kind. (I wrote this before I noticed your edit indicating that someone has stated a closed form solution for this, so apparently my intuition was wrong.)

At that stage you might ask yourself: is the whole formalism worth the trouble, or do you go for some Monte-Carlo simulation instead, where you construct uniformly distributed triples of points, check whether the circumcircle stays within the unit disk, and use the numeric evidence of that as your approximation. The choice probably depends on how much accuracy you need.

• Now he edited the question to make it meaningful, I have removed my downvote. Commented Apr 21, 2023 at 8:30