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If I pick a point within the circle about the origin of radius $R$, say $(r,\theta) = (0.5 R, \frac{\pi}{2})$, what is the average distance of all other points to that point?

Things which are interesting but not quite what I want:

  1. The average distance from the centre of a circle to any point within it is $\frac{2}{3}$ of the radius.

  2. If two random points are picked, the average distance is $\dfrac{128 R}{45\pi}$ (http://mathworld.wolfram.com/DiskLinePicking.html)

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  • $\begingroup$ A quick introduction to posting mathematical expressions may interest you. $\endgroup$ – hardmath Aug 8 '17 at 17:03
  • $\begingroup$ Note that by symmetry the "average distance" you want does not depend on the polar angle $\theta$. $\endgroup$ – hardmath Aug 8 '17 at 17:08
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    $\begingroup$ By "all other points" do you mean points in the disk -- the region surrounded by the circle -- or do you mean points on the circle, i.e. on the boundary of the disk, or do you mean something else? $\endgroup$ – Michael Hardy Aug 8 '17 at 17:10
  • $\begingroup$ In order to post an answer, you must first clarifies your question along the @MichaelHardy comment. $\endgroup$ – Felix Marin Aug 9 '17 at 2:36
  • $\begingroup$ Thanks hardmath, yes I suppose the function only depends on the radius. And yes, @Michael Hardy I mean all the points in the disk. Thanks to all for having a look. $\endgroup$ – MRMDP Aug 9 '17 at 17:16
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OK, I think the answer must be given by: $$ \int_{0}^{2\pi}\int_{0}^{R}r\sqrt{a^2+r^2-2ar\cos{\theta}} dr d{\theta} $$ Where $a$ is the distance of the point from the center, $R$ is the radius of the disc. Wolfram Alfa won't work this out and I haven't the inclination (or maybe ability) to do it myself.

I have done it numerically though, and the result is shown below (where $R=100$). As you can see when $a=0$, we get 66.66 ($\frac{2R}{3}$) as expected.

Within this range it is well approximated with a quadratic.

Average distance to a point within a disc

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