The average distance across a circle

Choose two points independently and uniformly on the circumference of a circle of radius \(1\).

What is the expected value—the average—of the straight-line distance between the two points?

Two points A and B on a unit circle, joined by a chord. A dashed radius from the center O to A is labeled 1.

The distance is measured along the chord joining the points, rather than along the circumference.

Explore the solution

The expected distance is \(4/\pi\approx1.273\).

Let \(\theta\in[0,\pi]\) be the smaller central angle between the two points. After fixing the first point, the second point’s angle relative to it is uniform on \([0,2\pi)\). Folding the two halves of this interval onto \([0,\pi]\) shows that \(\theta\) is uniform there, with density \(1/\pi\). The answer is the same whichever first point we fix, by rotational symmetry.

The two radii and the chord form an isosceles triangle. Bisecting it gives a right triangle with hypotenuse 1 and angle \(\theta/2\), so the chord length is

\[D=2\sin\!\left(\frac{\theta}{2}\right).\]

Averaging this length over the uniform angle gives

\[\begin{aligned} \mathbb E[D] &=\frac{1}{\pi}\int_0^\pi2\sin\!\left(\frac{\theta}{2}\right)\,d\theta\\ &=\frac{1}{\pi}\left[-4\cos\!\left(\frac{\theta}{2}\right)\right]_0^\pi\\ &=\boxed{\frac{4}{\pi}}. \end{aligned}\]

For a circle of radius \(R\), all distances scale by \(R\), and the same calculation gives \(4R/\pi\).