If we know the average distances between two points in a circle diameter 1 and the fact that all the circles are proportional to each other (pi is used as a constant), then is possible to find an equation to calculate it? Is it's so, then how would it be?
I have no idea how to approach, could we use it as pi or something like that?
Hope you can help me.
This is called disk line picking. For demonstration, let's use the unit circle. Essentially, to ensure a uniform distribution, we choose $x_i = \sqrt{r_i}\cos(\theta_i)$ and $y_i = \sqrt{r_i}\sin(\theta_i)$, and then we can calculate the integral $$\frac{\int_0^{2\pi}\int_0^1\int_0^{2\pi}\int_0^1 \sqrt{r_1+r_2-2\sqrt{r_1r_2}\cos(\theta_1-\theta_2)}\,\mathrm{d}r_1\,\mathrm{d}\theta_1\,\mathrm{d}r_2\,\mathrm{d}\theta_2}{\int_0^{2\pi}\int_0^1\int_0^{2\pi}\int_0^1 \mathrm{d}r_1\,\mathrm{d}\theta_1\,\mathrm{d}r_2\,\mathrm{d}\theta_2}$$
Edit: The transformation we choose is sort of explained on this page about disk point picking. Basically, if you uniformly randomly pick an $r\in (0, 1]$ and a $\theta\in [0, 2\pi)$, you'll get equal concentrations of points in each annulus, which you don't want (because an annulus further from the center will have the same number of points on average as the one closer to the center, even though the further one has a greater area). To control for this, we use the transformation described above, since annular area scales with $r^2$. We can calculate this explicitly by ensuring that $\mathrm{d}A = \mathrm{d}x\wedge \mathrm{d}y$ is constant in $r$. For $x = r\cos(\theta)$, $y = r\sin(\theta)$, we get $\mathrm{d}A = r\,\mathrm{d}r\,\mathrm{d}\theta$, but for $x = \sqrt{r}\cos(\theta)$, $y = \sqrt{r}\sin(\theta)$, we get $\mathrm{d}A = \frac{1}{2}\,\mathrm{d}r\,\mathrm{d}\theta$.
This problem (as well as a more general one for a compact convex set) has been solved in the paper "The average distance between two points" by B. Burgstaller and F. Pillichshammer.
The answer for two random points in the interior of a unit circle is: $$ \mbox{ the average distance } = {128\over45\pi}. $$
(See Example 2 on p.5 in the paper.) If the radius is $r$, the answer is $\displaystyle{128r\over45\pi}$.
As expected, this is less than $\displaystyle{4r\over\pi}$ (the answer for two random circumference points).
Back to the specific question
Suppose we know that, for a circle of diameter $=1$, the average distance between two random points is $a$; then for diameter $d$ the average distance is $da$. Likewise, if $a$ is the average distance for a unit circle (with radius $1$), then for radius $r$ the answer is $ra$.
(All circles are similar; so the answer can be obtained by changing the unit of measurement.)
Let $r$ be the radius of the circle. We will solve the problem for two random points on the circumference (the boundary of the circle).
If the two random points are separated by a random arc with central angle $x$, then the random distance $h=h(x)$ between the points is $$ h(x) = 2r\sin{x\over2} $$ and the average distance is $$ \bar h = {1\over\pi} \int_0^\pi 2r\sin{x\over2}\,dx = {4r\over\pi}. $$
If the circle is a ball in $\Bbb R^3$ \begin{equation} P_3 (r) = \frac{3 r^{2}}{R^3} - \frac{9 r^{3}}{4 R^4} + \frac{3 r^{5}}{16 R^6} \end{equation}
From this question,
Distances between randomly distributed points in a ball
which refers to an article for general formulas.
