POints in a circle connected as per a certain condition

Question

Suppose we chose $n$ points uniformly and randomly in a unit circle $\mathbf{S}$ centered at the origin $(0,0)$ and chose a number $r$,$0\leq r\leq1$.Connect two points by a line segment iff the product of their radial distances is less than $r$. Now given a point $Q$ at a radial distance $q$, how can we find the probability distribution of the number of points connected to it?I see a point with radial distance $u$ among the remaining is connected to $Q$ if the following holds: $$ u<\frac{r}{q}.$$ We may also note that the pdf of $u$ is given by $$ f(u)=2u,0\leq u\leq 1\\=0,\text{else} .$$Any help is greatly appreciated.Thanking you in advance.

Answer 1

Here is my attempt at answering my own question .Please point out the errors and mistakes if any.Given a point at a radial distance $q$ a point with a radial distance $u$ is connected to it if $$ uq<r$$ or,$$ u< \frac{r}{q}.$$There are now two cases $\textbf{Case 1}:$ $r<q $ In that case $u<1$
$\textbf{Case 2}:$ $r>q$ In that case $\frac{r}{q}>1$ Hence the probability that a point at at the radial distance $u$ is coonected to $u$ is given by $$ P(u<\frac{r}{q})=\begin{cases}\left(\frac{r}{q}\right)^2\,,r<q\\1\,&\text{else} \end{cases}.$$ In other words the probability thaat randomly chosen point will be connected to a point at a given radial location $q$ is given by $$ F(q)=\begin{cases} \left(\frac{r}{q}\right)^2\,,r<q\\ 1 &\text{else}\,\,\end{cases}$$.So it follows that the degree distribution of at radial distance $q$ is given by $$P(X=i)=\begin{cases} \binom{n-1}{i}(F(q))^{i}{\left(1-F(q)\right)^{n-1-i}}\,,r<q\\[2ex] n-1 &\text{else} \end{cases},$$,where $i=0,1,2,\cdots n-1$