# What is the average length of 2 points on a circle, with generalizations

I have earlier seen the question about finding the average length of two points and $n$ points inside the unit disk. But what about the more simple question, what happens if the points lie exactly on the circle?

I did some basic algebra, assume that the radius of the circle is $r$. without loss of generalization we can fix the point $x_1(-a,0)$ and then vary $x_2$ over all the possibilities $$P_2(2) = \frac{2 \int_{-a}^{a} \sqrt{2r(r+x)\,}\,}{2 \pi r^2} = \frac{16}{3\pi}$$ I also tested the results through matlab, generating $10^9$ points, and measuring the avreage distance. My results for a unit circle was $$D \approx 1.1864$$ Which unfortunately did not yield any result wheter in Wolfram, OEIS, or a inverse symbolic generator. This contradicts my arithmetic, meaning I should be wrong. The code snippet can be found below, nothing terribly exciting.

a = 1;
N = 10^8;

x1= a.*(2*rand(N,1)-1);
y1 = a.*(2*rand(N,1)-1);

x2= -a;
y2 = 0;

P = mean(sqrt((y2-y1).^2 + (x2-x1).^2));


What is the correct answer to $P_2(2)$ (average distance of two points on a circle)? Can one make any generalizations to $P_n(2)$ (average distance of two points on the the $n$-dimensional sphere? What about $P_n(m)$ or $P_2(m)$ ? Where $n$ is the dimension of the sphere, and $m$ is the number of points.

• Puzzling calculation, if the radius is $r$ the answer will be a constant times $r$. So we can take the radius to be $1$. There are several ways to choose the second point "at random," which need not give the same answer. The most natural (?) is to choose the angle uniformly distributed. Dec 18 '13 at 15:14
• @AxelKemper: It is (here) early in the morning, pre-coffee. But with angle uniformly distributed, I do not see an elliptic integral, get $\frac{4}{\pi}$. But OP chooses "$x$" uniformly distributed. Dec 18 '13 at 15:45

I don't know how you arrived at your formula for "$P_2(2)$", whatever is meant by this.

At any rate you may assume your circle of radius $1$, the first point $z_1$ as $(1,0)$ and the second point $z_2$ as $(\cos\phi,\sin\phi)$ with $\phi$ equidistributed on $[0,\pi]$. Then $|z_2-z_1|=2\sin{\phi\over2}$ and therefore $${\mathbb E}\bigl[|z_1-z_2|\bigr]={1\over\pi}\int_0^\pi 2\sin{\phi\over2}\ d\phi={4\over\pi}\doteq1.273\ .$$

Your Matlab code generates points in a square, not on a circle. You should change it into something like this:

a = 1;
N = 10^8;

theta = 2*pi*rand(N,1);
x1 = a.*cos(theta);
y1 = a.*sin(theta);

x2 = -a;
y2 = 0;

P = mean(sqrt((y2-y1).^2 + (x2-x1).^2));


This gives about 1.274, which doesn't agree with 16/3/pi.