Distribution of line intersections on the plane

Question

Given two circles of radius $r$ and $R$ and $n$-lines in the plane $\mathbb{R}^2$. Given, that none of the lines are parallel, these lines have at most $\frac{(n^2-n)}{2}$-intersection points.

Further, we assume that all lines have to intersect the circle of radius $r$, my question is how many of the intersections lie outside of the circle of radius $r$ but within the circle of radius $R$ in dependence of $R$?

E.g. in the picture below all lines intersect the orange circle of radius $r$. 3 intersection points (black) lie within the orange circle, 3 (red) in the ricle of radius $R$ and one (green) outside of both circles.

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A mathematical description: We assume we have $n$-lines $L_n$ given by two points in the plane $(x_i^1,y_i^2),(x_i^3,y_i^4) \quad i=1,...,n$. We assume that the first point is drawn uniformly from the small circle $(x^1_i,y^2_i) \sim U(\sqrt(x^2+y^2)<r)$ to ensure the condition that all lines intersect the small circle with radius $r$. The second point $(x^3_i,y^4_i)$ is uniformly drawn from $U(\sqrt(x^2+y^2)<R)$. The intersection of two lines is given by

$$\begin{array}{l}P_{x}=\frac{\left(x^1_{1} y^4_{1}-y^2_{1} x^3_{1}\right)\left(x^2_{2}-x^2_{3}\right)-\left(x^1_{1}-x^3_{1}\right)\left(x^2_{1} y^4_{2}-y^2_{2} x^3_{2}\right)}{\left(x^1_{1}-x^3_{1}\right)\left(y^2_{2}-y^4_{2}\right)-\left(y^2_{1}-y^4_{1}\right)\left(x^1_{2}-x^3_{2}\right)} \\ P_{y}=\frac{\left(x^1_{1} y^4_{1}-y^2_{1} x^3_{1}\right)\left(y^2_{2}-y^4_{2}\right)-\left(y^2_{1}-y^4_{1}\right)\left(x^1_{2} y^4_{2}-y^2_{2} x^3_{2}\right)}{\left(x^1_{1}-x^3_{1}\right)\left(y^2_{2}-y^4_{2}\right)-\left(y^2_{1}-y^4_{1}\right)\left(x^1_{2}-x^3_{2}\right)}\end{array}$$

now we need to calculate the expectation value for $(P_x,P_y)$ given the above restrictions on how $(x_1^1,y_2^2),(x_2^3,y_2^4)$ is drawn and generalize the result to $n$ lines. But here I am stuck.

Also, any hint at existing literature dealing with similar problems would be much appreciated.

Answer 1

Too long for a comment:

I suspect you are going to find this difficult.

Clearly with $n$ lines, there will be $\frac{n(n-1)}2$ intersections almost surely. The distribution of their intersections' distances from the common centre of the circles (you have implicitly suggested they are concentric) will depend on how you select them (see Bertrand's paradox - I suspect you want the lines' intersections with the $r$ circle to be uniformly distributed on its circumference but you have not said this), and will not be independent of each other, though may be close to independent for large $n$. You could consider the expected number of intersections within the $R$ circle, so $\frac{n(n-1)}2$ times the probability of an individual intersection being within that circle. You may as well take $r=1$ and rescale $R$.

This simulation using R indicates the many of the intersection distances are close to $1$ (and clearly $\frac13$ are less than $1$) but this is a heavy tailed distribution - I doubt the second moment is finite and I am not sure about the first moment.

intersect <- function(x,y){
  c((y[3]*x[2]*x[4]-y[1]*x[2]*x[4]-y[3]*x[1]*x[4]+y[2]*x[1]*x[4]-
     y[4]*x[2]*x[3]+y[1]*x[2]*x[3]+y[4]*x[1]*x[3]-y[2]*x[1]*x[3])/
    (y[2]*x[4]-y[1]*x[4]-y[2]*x[3]+y[1]*x[3]-
     y[4]*x[2]+y[3]*x[2]+y[4]*x[1]-y[3]*x[1]),
    (x[3]*y[2]*y[4]-x[1]*y[2]*y[4]-x[3]*y[1]*y[4]+x[2]*y[1]*y[4]-
     x[4]*y[2]*y[3]+x[1]*y[2]*y[3]+x[4]*y[1]*y[3]-x[2]*y[1]*y[3])/
    (x[2]*y[4]-x[1]*y[4]-x[2]*y[3]+x[1]*y[3]-
     x[4]*y[2]+x[3]*y[2]+x[4]*y[1]-x[3]*y[1]))
   } 

rintersect <- function(){
  theta <- runif(4)*2*pi
  x <- cos(theta)
  y <- sin(theta)
  point <- intersect(x,y)
  sqrt(point[1]^2 + point[2]^2)
  }

set.seed(2023)
sims <- replicate(10^6, rintersect())   
mean(sims<1) # expected to be 1/3
# 0.332991

giving an empirical density for $\frac Rr$ which looks like

plot(density(sims, from=0, to=3))
abline(v=1, col="red")

and an empirical cumulative distribution function which looks like

plot.ecdf(sims, xlim=c(0,20))
abline(v=1, col="red")
abline(h=1/3, col="red")

Your example diagram actually has ten points of intersection of the five lines though some are not visible: $4$ inside the orange circle, $3$ between the orange circle and the blue circle with twice the radius, and $3$ outside the blue circle. The probabilities are about $0.33$, $0.40$ and $0.27$ and so the expected numbers when $n=5$ about $3.3$, $4.0$ and $2.7$, not too far away.