intersection of Voronoi cell and Circle My question is about overlapping a random Voronoi cell and a circle.
Suppose there are some Poisson Voronoi cells generated by a homogeneous Poisson Point Process with density λ and Voronoi partitioning. For an arbitrary cell region, consider a circle with fixed radius r centred at the seed (or site) of the cell, what's the average perimeter of the intersection of the circle and Voronoi cell region?
I can see that the overlapped area might have a mixture of Voronoi cell & circular boundary or purely Voronoi or purely circular depending on the realization of the Voronoi cell. Another thing I found is that the average boundary of a Poisson Voronoi cell is $\dfrac{4}{\sqrt{\lambda}}$, which should give us an upper bound of the perimeter.
Any hint on how to approach this problem? Much appreciated.
 A: Condition upon the site we are interested in being the origin (that is a somewhat sloppy phrase, of course, but not more sloppy than "arbitrary cell region"; once you make sense of the latter, you can make the corresponding sense of the former). The rest of the particles under this conditioning still form the Poisson process of intensity $\lambda$. 
Now consider the part of the perimeter lying in a small (actually infinitesimal) angle of opening $\alpha$ around the positive $x$-semi-axis. By the rotational symmetry, the total expectation of the perimeter is $2\pi/\alpha$ times the expectation of this part. The point $(t,0)$ is the intersection point of the positive $x$-semi-axis and the boundary of the Voronioi cell if the circle centered at this point of radius $t$ contains no Poisson point inside but has a point on the boundary. The probability that it will happen for $t\in[\rho,\rho+d\rho)$ is just $-d(e^{-\lambda\pi\rho^2})$. Now, assuming that, the distribution density of the random angle $\theta$ the direction to the corresponding Poisson point from the point $(\rho,0)$ makes with the positive $x$-semi-axis is proportional to the width of the corresponding crescent between the disks $D((\rho+d\rho,0),\rho+d\rho)$ and $D((\rho,0),\rho)$ as seen from $(\rho,0)$, i.e., to $(1+\cos\theta)d\rho$. The angle the boundary of the cell makes with the positive $x$-semi-axis is twice less, i.e., $\theta/2$. Hence the (conditionally) expected length is
$$
\rho\alpha\frac 1\pi\int_0^\pi\frac{1+\cos\theta}{\cos\theta/2}\,d\theta=
\rho\alpha\frac 1\pi\int_0^\pi 2\cos\frac\theta2\,d\theta=\frac 4\pi\alpha\rho\,.
$$
Now this works for $\rho<r$. For $\rho>r$, the circle takes over giving the length $r\alpha$. Integrating, etc., we get the expectation in question equal to
$$
2\pi\left[\frac 4\pi\int_0^r\rho(-de^{-\lambda\pi\rho^2})+re^{-\lambda\pi r^2}\right]=8\int_0^re^{-\lambda\pi\rho^2}\,d\rho+(2\pi-8)re^{-\lambda\pi r^2}\,.
$$
It is ugly, as promised, but not excessively ugly. You can rewrite it in a more beautiful way using the $\operatorname{Erf()}$ error function, which would allow you to compute the numerical values using any standard software, of course, but this is what it is. Note that this simple technique generalizes to the intersection with any convex shape and any dimension, but you'll get an even uglier multiple integral in polar coordinates as an answer.
Edit 1 Here is the picture showing the piece of the boundary of the Voronoi cell (red) near the point $(t,0)$. The point $p$ is another Poisson point at the same distance from $(t,0)$ as the seed $(0,0)$. The rest should be self-explanatory.
 
This explains $\cos\frac \theta 2$ in the denominator. I'll add one more picture explaining $1+\cos\theta$ in the numerator a bit later :-)
Edit 2. Here is the second picture. In order for the boundary of the Voronoi cell to intersect the positive $x$-semi-axis between $r$ and $r+dr$, the yellow circle should be empty and the blue crescent should contain a Poisson point. That point (conditioned upon its existence) has to be uniformly distributed in the blue crescent according to the area measure, which we are trying to write in polar coordinates. The red sides of the red-green triangle are $dr$ and $r+dr$, the (outer) angle between them is $\approx\theta$, so the green side is about $r+dr+dr\cos\theta$. The length of the green side within the yellow circle is $r$, leaving the infinitesimal crescent width at $dr(1+\cos\theta)$ (which should be multiplied by $r\,d\theta$ to get the full area element).

