## Picking Random Points on a Circle (Not a Disk)

How do you select points at random on a circle?  By “circle” I mean the outside of a disk, not its interior.  In this post I’m going to discuss two methods: (1) Selecting an angle at random on $[0, 2\pi]$ and taking the point that corresponds to that angle, and (2) Selecting an x-coordinate at random and taking the point that corresponds to that x-value.  (Actually, there are two possible points in Case (2), but we can subsequently choose either of those with a 50% probability.)  My main purpose is to show that the second method does not actually select points at random on the circle; instead, that method chooses points more likely to be near the top and bottom of the circle than near the sides.

First, we have to ask what it means, in a precise mathematical sense, to pick points at random on a circle.  I’m going to take this to mean the following:

A process that selects a point at random on a circle is one such that the probability of picking a point from an arc of the circle is proportional to the length of that arc.

For example, there should be a probability of 1/4 of picking a point from the upper-right quadrant of the circle, a probability of $1/2\pi$ of picking a point from an arc of length 1, and in general a probability of $L/2\pi$ of picking a point from an arc of length L.

For simplicity’s sake, let’s use a unit circle.  Suppose we pick an angle at random from $[0, 2\pi]$ and take the corresponding point on the circle.  The probability of selecting an angle from a particular interval of length L is $L/2\pi$.  Since the length of an arc of the unit circle is just $\theta$, where $\theta$ is the angle subtended by that arc, the probability of this process selecting a point from an arc of length L is also $L/2\pi$.  Thus selecting an angle at random and taking the point that corresponds to that angle is a process that selects a random point on the circle.

Why doesn’t selecting an x-coordinate do the same thing?  Let’s focus on just the upper-right quarter of the circle and select an x value at random from $[0,1]$.  Then we take the point $(x, \sqrt{1-x^2})$ on the circle.  Since x is related to $\theta$ by the relationship $x = \cos \theta$, and $\cos \theta$ is decreasing on $[0, \pi/2]$, the probability of the point $(x, \sqrt{1-x^2})$ ending up in the range $[0, \pi/4]$ of $\theta$ values is the probability that x is chosen from the interval $[ \cos(\pi/4), \cos(0)] = [\sqrt{2}/2, 1]$.  This probability is not $1/2$ (as it would be if this process selected a point at random on the unit quarter-circle) but $1 - \sqrt{2}/{2} \approx 0.292$.  Thus choosing a random x from $[0,1]$ and taking the corresponding point on the unit quarter circle tends to produce more points near the top of the circle than near the side.

Generalizing from the example, the probability that the x-coordinate procedure produces a point on the circle with an angle in the range $[0, \theta]$ is $1 - \cos \theta$.  This means that the probability of the x-coordinate procedure yielding a value in a small range of width $\Delta \theta$ around a fixed $\theta$ value is $\frac{d}{d \theta} (1 - \cos \theta) = \sin \theta$ times $\Delta \theta$, or $\sin \theta \Delta \theta$.  Since the sine function increases on $[0, \pi/2]$, the closer $\theta$ is to $\pi/2$, the more likely it is that the x-coordinate procedure will produce a value near $\theta$.