Probability describing how many adjacent hexagons overlap a circle of a given radius In a simulated landscape comprised of tiled hexagons of a uniform size, if I were to drop a circle in this landscape at a random location, how could I determine a probability distribution function which describes the  many hexagons would overlap (including partial overlaps) the randomly placed circle.

The example image shows 22 hexagons at least partially overlapping the circle.  So the question would be what is the probability that I'd get this value (or alternate values) if the circle was randomly placed on this landscape?
For a bit of context here, I'm interested to see if I can get some probability distribution function for species distribution mapping which accurately represent the life history of a particular species.  In this case, I'm interested in a probability distribution function describing how many bird territories are within an observers sampling area.
I currently just use simulations to get a distribution resembling this, but I'm sure there's a cleaner, more efficient way to derive this distribution from geometric proofs.  My question is somewhat similar to this.
Note:  My math is pretty amateur, so if please let me know if my question is unclear.
Edit 1:
Perhaps to make this easier (maybe?), I'm happy with an approximation to the solution, so perhaps using slightly overlapping circles rather than hexagons makes the problem more feasible?
 A: The answer depends on the radius of the circle. The case of the hexagons is much more complicated than the case of circles. So we assume that instead of the hexagons you have circles of radius $r$ centered at
the points of the grid $mU+nV$ for $n,m\in\Bbb{Z}$, $U=(1,0)$ and $V=\left(\frac 12,\frac{\sqrt{3}}{2}\right)$,
with $\frac 12\le r \le \frac{\sqrt{3}}{3}$. 
If you want to know the probability distribution function which describes how many small circles would overlap a randomly placed circle of radius $R$ and center $P$, you have to draw circles of radius $r+R$ around each nearby point in the grid and the point $P$ in the plane is inside a  certain number $N=N(P)$ of these big circles. The circle of radius $R$ would overlap with $N$ small circles. 
The probability you are searching is the proportion of the area of the plane where all points 
have $N(P)=N$. By symmetry you can restrict the computations to the area inside the triangle $A$ formed by 
$$
(0,0),\left(\frac 12,0\right)\quad\text{and}\quad \left(\frac 12,\frac{\sqrt{3}}{6}\right).
$$
Then the circle of radius $R$ will cut the triangle into pieces, and all the points in each piece would be all inside the same set  of big circles with radius $R+r$. So for all points in one piece the value of $N(P)$ is the same. The area of all the pieces corresponding to a fixed $N$, divided by the area of the triangle, is the probability that a randomly placed circle will overlap $N$ small circles.
So the computation reduces to the area of pieces of a triangle cut by circular arcs.
For example consider the case where $r+R=2$. Then the triangle $A$ is inside the circles centered at $mU+nV$ for $(m,n)\in\{(-1,0),(0,0),(1,0),(2,0),(0,-1),(1,-1),(2,-1),(-1,1),(0,1),(1,1),(-1,2),(0,2)\}$ and overlaps partially the circles centered at
$mU+nV$ for $(m,n)\in\{(-2,1),(-1,-1),(1,-2),(2,-2)\}$. So each randomly placed circle overlaps (partially or completely) at least 12 small circles and at most 16.
Now comes the clumsy part: You have to cut the triangle into pieces and compute the area of each piece. The equations of the four critical circles are:
$$
\left(x +\frac32\right)^2 + \left(y - \frac{\sqrt{3}}2\right)^2 = 4,\quad \left(x + \frac32\right)^2 + \left(y + \frac{\sqrt{3}}2\right)^2 = 4,
$$
$$
x^2 + (y + \sqrt{3})^2 = 4,\quad (x -1)^2 + (y + \sqrt{3})^2 = 4.
$$
The triangle is cut into 7 pieces, two pieces correspond to $N=14$, two pieces correspond to $N=15$, and one piece each correspond to 12,13 and 16 respectively.
Now you have to do basic geometry on the plane or use calculus.
 For example, the area of the piece corresponding to $N=16$ is
$$
R=\int_0^{a_0}\left(-\sqrt{3}+\sqrt{4-(x-1)^2}\right)dx+\int_{a_0}^{b_0}\left(-\sqrt{3}/2+\sqrt{4-(x+3/2)^2}\right)dx\sim 0.01882,
$$
where $a_0=\frac{1}{28}(-7+3\sqrt{21})$ and $b_0=(\sqrt{13}-3)/2$.
Since the area of $A$ is $\frac{\sqrt{3}}{24}\sim 0.0721688$, we obtain
that the probability of overlapping(partially or completely) 16 circles is 
$$
P(16)=\frac{Area\ of\ the\ piece\ corresponding\ to\ 16}{Area\ of\ A}\sim 0.26077822.
$$
