Let me explain you my reasoning and next help me to find if there is something wrong.
I state the problem in polar, not euclidian, coordinates.
Consider an infinitesimal element enclosed between circles of radii r and r+dr and vector radii θ and θ+dθ.
Its surface is r*dr*dθ.
Now, are you agree with me if I say that if the N points within the circle are uniformly distributed, then each of them carries a mass equal to Pi*R^2/N.
To argue, lets draw a parallel with the Monte-Carlo summation formula :
- Consider a square (or any other figure a uniform sampling is easy to obtain)os area A which contains the disc and sample M points in this shape : the surface of the disc is just the ratio N/M*A where N is the number of points out of the M that fall within the disc.
This simple formula means that every points have exactly the same mass, saying 1, (if it wasn't the case the good formula should account for each specific weight wn and would a little bit more complicated).
The idea behind is those of an "unweighted sample", or in other words "fair apportionment of the mass".
Now, suppose the element [r, r+dr]x[θ, θ+dθ] contains K(r, dr, dθ) points
Consider now the domain [2*r, 2*r+dr]x[θ, θ+dθ] : its area (mass) is twice the mass of the previous one and the number of points it contains is K(2*r, dr, dθ).
By "fair apportionment of the mass", each point having the same mass one must have K(2*r, dr, dθ) = 2*K(2r, dr, dθ).
The factor 2 can be replaced by any other value (provided we are still within the disc of course) ; which shows that the radial density of the points is proportional to r.
I take it that the pdf of the radius is of the form C*r where C is some suitable normalization constant.
Equating int(C*r, r=0..R) to 1 I find then C=2/R^2.
If the pdf of r is 2*r/R^2, then r is distributed according a triangular law (writen in Maple Triangular(0, R, R).
I disagree with you when you say "its a kind of statistical intuition".
But I would admit without shame that my reasoning is wrong as soon as someone (you ?) will have point out where its failure is
So, please feel free to defuse my arguments. One always learn from ones mistakes
Hope to read you soon.