Assuming I observe, in a unit square, $n_1$ circles of area $A_1$ (non-overlapping amongst themselves) and $n_2$ circles of area $A_2$ (again, non-overlapping) and that each of the centres is uniformly distributed across the square (well, I guess that can't be precisely true given the non-overlapping constraint), is there an expression for the proportion of the first circles that overlap with at least one of the second type of circles.

The actual situation is observed cancer cells and pores on a microscope slide, to which the above seems a good approximation.

In the 1D case, I'd like to think that for each line segment (1D equivalent of a circle) of the first group, there's an interval of size $A_1+ A_2$ before the end of each segment of the second type to place the start of the interval so that they would overlap, and so $n_2(A_1+A_2)$ of the unit interval would result in an overlap. That seems to be true for the first segment of the first group, but it won't necessarily hold for further segments as the non-overlapping criterion breaks the independence, so I'm probably looking at this (and the 2D case) from the wrong angle.

Updated thoughts

Maybe this is the way to look at it? There's a total area $n_1A_1$ covered by type 1, and $n_2A_2$ covered by type 2, out of a total area of 1. So an expected area of $n_1A_1n_2A_2$ is covered by both, and divide this by $A_1$ to give the number of type 1's that would be needed to cover that area, and divide by $n_1$ to get the proportion of type 1's that are in the double-covered area. So if 7% of a slide's area is covered by spores, then 7% of cells are hit by spores?

  • $\begingroup$ Are the middles of the circles within the square, or the whole circles? $\endgroup$
    – Angelorf
    Mar 5, 2014 at 11:30
  • $\begingroup$ The circles are going to be small compared to the square (and also reasonably sparse, not taking more than 10% of the available area), so I'm hoping it doesn't make too much difference - but to be accurate, this is a square subregion so it's possible that even the centre of a circle is outside (but no further than the circle's radius away from) the boundary. $\endgroup$ Mar 5, 2014 at 11:38
  • $\begingroup$ Are the $n$'s generally big? $\endgroup$
    – Angelorf
    Mar 5, 2014 at 11:41
  • $\begingroup$ 10% .. is that the total area of the circles, or the area of each? $\endgroup$
    – Angelorf
    Mar 5, 2014 at 11:42
  • $\begingroup$ That's of the total area. So no more than 10% of the square is covered by circles of type 1, and no more than 10% is covered by type 2 circles. $\endgroup$ Mar 5, 2014 at 11:57

1 Answer 1


An easy procedure would be to calculate the chance that two given circles $c_1$ (in the set $C_1$, of area $A_1$) and $c_2$ are overlapping and use the number of combinations of a circle of each group resulting in a number $N$ overlaps.

I'm assuming that you mean that each circle in the first group has area $A_1$ and each in the second $A_2$.

My procedure does simplify some issues concerning borders and overlap of multiple circles. I assume the circles are relatively small and few.

The chance of a circle $c_1$ of area $A_1$ to have overlap with another circle $c_2$ of area $A_2$ within an area of $1$, is the area given by the possible positions of the centre of $c_2$ causing an overlap with some given $c_1$. This area forms a circle with a radius equal to the sum of the radii of $c_1$ and $c_2$. Since $A = \pi r^2$, $r = \sqrt{A/\pi}$ and so the area under consideration is $A_{\text{overlap}} = \pi (\sqrt{A_1/\pi} \cdot \sqrt{A_2/\pi})^2$. The ratio of this area to the total area gives the chance $P_{\text{overlap}}(c_1, c_2) = A_{\text{overlap}} / A_{\text{total}}$.

The expected number of overlaps is $E(N) = \sum\limits_{n=0}^{M} nP(N=n) $, where $M=\min(n_1,n_2)$ is the maximum number of overlaps assuming that no circle is overlapping multiple others. When we further ignore the fact that circles in $C_2$ that don't overlap circles of $C_1$ shouldn't overlap amongst themselves, we can say that $P(N=n) = {n_1\choose n}{n_2\choose n}n! (P_{\text{overlap}})^n (P_{¬ \text{overlap}})^{n_2-n} $. Here ${n_1\choose n}{n_2\choose n}n!$ gives the number of possibilities for $n$ pairwise combinations from groups $n_1$ and $n_2$

When assuming that each two circles of $C_1$ are at a distance between them that leaves a space for a non-overlapping circle of $C_2$ between them (so we assume that for each circle in $C_1$ its area of overlap doesn't overlap with the area of overlap with any other area of overlap of a circle in $C_1$), then the chance of a circle $c_2$ of area $A_2$ having no overlap with any circle of $C_1$ is $P_{¬ \text{overlap}} = (A_{\text{total}}- A_{\text{overlap}}*n_1)/A_{\text{total}}$

Since $A_{\text{total}} = 1$, we then have $P(N=n) = {n_1\choose n}{n_2\choose n}n! (A_{\text{overlap}})^n (1- A_{\text{overlap}}*n_1)^{n_2-n}$, where $A_{\text{overlap}} = \pi (\sqrt{A_1/\pi} \cdot \sqrt{A_2/\pi})^2$

There have been a lot of simplifications and/or assumptions along the way, but I think they are not so bad given that the circles are small and sparse.

I know this answer isn't very formal and precise, but I hope it helped.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.