# Average distance between two points in a circular disk

How can I find an average distance between two points lying inside a circular disk of a certain radius?

I wonder if there is any other way except of using a Monte Carlo method?

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See the answer to this question. The expected distance is $$d= {128 r\over 45\pi}.$$

Here is another demonstration of this result.

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Your answer seem to be wrong. Using Monte Carlo simulations (see this script) I got another answer $d=0.72r$, and $\frac{128}{45\pi}\approx0.91$. – installero Apr 24 '12 at 6:05
You're probably making the mistake decribed here anderswallin.net/2009/05/… – Gebb Apr 24 '12 at 8:44
Exactly! Thank you! – installero Apr 26 '12 at 15:12

Let $D$ be the disk, and let $$M = \iint\limits_{(x_0,y_0)\in D} ~~\iint\limits_{(x,y) \in D}dxdydx_0dy_0.$$

Then the quantity you are looking for should be given by $$\frac{1}{M}\iint\limits_{(x_0,y_0)\in D} ~~\iint\limits_{(x,y) \in D} \sqrt{(x-x_0)^2+(y-y_0)^2}dxdydx_0dy_0$$ where a suitable change to polar coordinates in both double integrals would probably be helpful.

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Unfortunately, this method seems to lead me to an unsolvable finite integrals. I would be very thankful if you can help me getting to a final answer. – installero Apr 23 '12 at 13:39

With the probability density you can find the average of the distance, or the distance squared, or the variance of the distance, or whatever you want. The probability density for the distance $l$ between points in a circle of radius $r$ is given by Ricardo García-Pelayo 2005 J. Phys. A: Math. Gen. 38 3475 as $$p(l)=\frac{4l}{\pi r^2}\arccos \frac{l}{2r}-\frac{2l^2}{\pi r^4}\sqrt{r^2-\frac{l^2}{4}}$$

Then the average distance is given by $$\int_0^{2r}lp(l)dl=\frac{128r}{45\pi}.$$ The average of the distance squared is given by $\int_0^{2r}l^2p(l)dl=r^2.$ Thus the standard deviation is given by $$\sqrt{\int_0^{2r}(l-\frac{128r}{45\pi})^2p(l)dl}=\sqrt{r^2-\left( \frac{128r}{45\pi} \right)^2}=\frac{r\sqrt{2025\pi^2-2^{14}}}{45\pi}$$

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