Two points are chosen at random within a circle of radius R. What is the probability that the distance between them is less than D?

I can solve the problem for the special case $R=D$ as follows:

For simplicity assume $R=D=1$

Let $O$ be the centre of the original circle of radius $R=1$. Let $A$ and $B$ be the two random points and let $d$ be the distance between them.

Suppose the distance $OA$ is $r$. Then we draw another circle of radius $D=1$, centred on $A$. Then the conditional probability $P(d<1|OA=r)$ is the area of the intersection of the two circles divided by the area of the original circle.

Let $C$ and $D$ be the points of intersection of the two circles. Now $CD$ is a line of symmetry and the area of intersection of the two circles is twice the area of the smaller segment bounded by $CD$ and the original (or equivalently the other) circle.

Intersection area = $2(\frac{1}{2}(1^2)(\theta - sin(\theta)))=\theta - sin(\theta)$, where $\theta=\angle COD$

Therefore $P(d<1|OA=r)=\frac{\theta - sin(\theta)}{\pi(1)^2}=\frac{\theta - sin(\theta)}{\pi}$

Where $cos(\frac{\theta}{2})=\frac{r}{2}$

Rearranging gives $r=cos(\frac{1}{2}\theta)$

Hence $\frac{dr}{d\theta}=-sin(\frac{1}{2}\theta)$

Now we are ready to do the integral:

$P(d<1)=\int_0^1(\frac{\theta - sin(\theta)}{\pi})\frac{2\pi r}{\pi(1)^2}dr$

$=\frac{2}{\pi}\int_0^1(\theta - sin(\theta))(r)dr$

$=\frac{2}{\pi}\int_\pi^{\frac{2\pi}{3}}(\theta - sin(\theta))(r\frac{dr}{d\theta})d\theta$

$=\frac{2}{\pi}\int_\pi^{\frac{2\pi}{3}}(\theta - sin(\theta))(-2cos(\frac{1}{2}\theta)sin(\frac{1}{2}\theta))d\theta$

$=\frac{2}{\pi}\int_\pi^{\frac{2\pi}{3}}(\theta - sin(\theta))(-sin(\theta))d\theta$

$=\frac{2}{\pi}\int_{\frac{2\pi}{3}}^\pi(\theta sin(\theta) - sin^2\theta)d\theta$

Computing this integral (us integration by parts for the first term and the half-angle formula for the second) gives a probability of $1-\frac{3\sqrt{3}}{4\pi}$ ($0.5865$ to 4dp), a result which I have verified to 2dp using a random number simulation.

When I tried to use a similar method for the general case, where D is not equal to R, I found that the equations for $r$ in terms of $\theta$ was quadratic and the resulting integral would have been extremely complicated. I thought perhaps a different approach might be called for.

  • $\begingroup$ I do not see any way to avoid elliptic integrals in the general case. $\endgroup$ – Jack D'Aurizio Aug 31 '18 at 19:44

According to MathWorld, the probability density for the distance $s$ between two points uniformly randomly chosen in a disk of radius $R$ is

$$ f(s)=\frac{4s}{\pi R^2}\arccos\frac s{2R}-\frac{2s^2}{\pi R^3}\sqrt{1-\left(\frac s{2R}\right)^2}\;. $$

Thus the probability for the distance to be at most $s$ is

\begin{eqnarray*} F(s)&=&\int_0^tf(t)\mathrm dt \\ &=& 1-\frac2\pi\left(1-\frac {s^2}{R^2}\right)\arccos\frac s{2R}-\frac s{\pi R}\left(1+\frac{s^2}{2R^2}\right)\sqrt{1-\left(\frac s{2R}\right)^2}\;, \end{eqnarray*}

in agreement with your result for $s=R$.

Here's a plot for $R=1$.


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