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Find⠀the⠀average⠀distance⠀of two⠀randomly-selected⠀points⠀in⠀a⠀circle⠀of⠀Radius $=R$.

This⠀is⠀my⠀approach: Consider a function $I(R)$ that is given an input (the radius $R$ of the circle) and outputs the average distance between two randomly-selected points inside that circle. The size of the radius seems like it would only scale the answer, so we can say that $I(R)=R\cdot I(1)$. This reduces the problem to merely considering the case when $R=1$.

One’s first instinct might be to consider two random variables, $r\in [0,1]$ and $\theta \in [0,2\pi]$, and considering $x=r\cos\theta$ and $y=r\sin\theta$, but a closer inspection shows us that these won’t correspond to random points, and the points would cluster more around the center of the circle than they would closer to the edges (notice to that $dA=r\cdot dr\cdot d\theta$, not just "⠀$dr\cdot d\theta$ "). The correct transformation is $x=r\sqrt{\cos\theta}$ and $y=r\sqrt{\sin\theta}$ (when I first considered this problem myself, I understood that the first approach wouldn’t work, and the points would cluster around the center, but even now I couldn’t exactly tell you why the correct transformation requires taking precisely the $\sqrt{r}$ ; this is evidence for the question’s conceptual difficulty).

Considering this transformation and integrating each variable with respect to its PDF, taking (without loss of generality) the first point to be $(r_1,0)$ and the second to be $(r_2,θ)$, and noticing that $\theta$ only needs to range from $0$ to $\pi$ yields the integral⠀:

\begin{align}I(1)&=\frac{\int_{0}^{1}\int_{0}^{1}\int_{0}^{\pi}\sqrt{(\sqrt{r_1}-\sqrt{r_2}\cos\theta)^2+(\sqrt{r_2}\sin\theta)^2}⠀d\theta dr_1 dr_2}{\int_{0}^{1}\int_{0}^{1}\int_{0}^{\pi}d\theta dr_1 dr_2}\\&=\frac{1}{\pi}\int_{0}^{1}\int_{0}^{1}\int_{0}^{\pi}\sqrt{r_1+r_2+\sqrt{r_1r_2}\cos\theta}⠀d\theta dr_1 dr_2 \end{align}

It took me 2hr to type the question and 1hr to type this integral. This took me really Long, So without showing the steps to anti-differentiate, This equals to : $⠀I(R)=\frac{128}{45\pi}R $ Is my answer is correct?⠀

Kindly provide⠀your⠀solution⠀too.

user400188
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  • I know that its a bit Lengthy , SORRY for that ~ –  Mar 23 '21 at 08:57
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    kindly edit to use a more informative title. – Siong Thye Goh Mar 23 '21 at 09:04
  • Is it ok if I modify the formatting for you so the spacing is consistent with the rest of the site? I see you took a long time writing it up and I don't want to spoil it, however it will be hard to read for some viewers in its current state. (formatting aside, it seems like a very good question). – user400188 Mar 23 '21 at 09:06
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    Thanks @user400188 –  Mar 23 '21 at 09:10
  • You are right: see here – Jean Marie Mar 23 '21 at 09:16
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    "Randomly" is not well-defined here. It will depend on which probability distribution on the circle you chose. Also, this is no differential geometry. – Didier Mar 23 '21 at 09:21
  • https://math.stackexchange.com/questions/135766/average-distance-between-two-points-in-a-circular-disk – Math Lover Mar 23 '21 at 09:55
  • Another link: https://math.stackexchange.com/questions/2946673/expected-distance-between-two-points-inside-unit-circle – Math Lover Mar 23 '21 at 09:56

1 Answers1

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The desired average is obtained by a quadruple integral in Cartesian coordinates.

$$\frac1{\pi^2}\int_{x^2+y^2\le1,\\u^2+v^2\le1}\sqrt{(x-u)^2+(y-v)^2}dx\,dy\,du\,dv$$

Switching to polar and using rotational symmetry, $$\int_0^{2\pi}\int_0^1\int_0^{2\pi}\int_0^1\sqrt{(r\cos\theta-s\cos\phi)^2+(r\sin\theta-s\sin\phi)^2}\,rdr\,d\theta\,s ds \,d\phi$$ $$=2\pi\int_0^1\int_0^{2\pi}\int_0^1\sqrt{r^2-2rs\cos\theta+s^2}\,rdr\,d\theta\,s ds.$$

From here, computation seems nightmarish. Integration on $\theta$ yields a complete elliptic function. Integration on $r$ or $s$ is doable but leads to relatively complex expressions.