# How to find the intersection of union of two circle groups

I have two groups of circles. S1 is the union of the first group and S2 is the union of the second group of circles. I know center and radius of all circles. I have to find the equation for the intersection of S1 and S2. Is there anyway to do it?

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For the sake of searchability, "circle group" has a different and precise mathematical definition (en.wikipedia.org/wiki/Circle_group) so it might be better to say something like "collection of circles." – Qiaochu Yuan Aug 3 '12 at 21:08
What do you mean by the "equation" for the intersection? The area? – mjqxxxx Aug 3 '12 at 21:12
I mean any mathematical representation of the area that i am looking for. – kotoll Aug 3 '12 at 21:20
The description you gave (the intersection of the respective unions of two known collections of circles) is already a mathematical representation of a subset of the plane. Whether there is another, more useful, representation depends on what you want to do with it. If you want to test whether a given point belongs to the intersection, that description is probably fine. If you want the perimeter or the area or the number of disjoint closed components, then not so much. – mjqxxxx Aug 3 '12 at 22:01

## 1 Answer

Intersection of unions is the union of intersections (intersection is distributive over union): $$\left( \bigcup_{c\in \mathcal S_1}c\right)\cap \left(\bigcup_{c\in \mathcal S_2}c\right)=\bigcup_{c_1\in \mathcal S_1,c_2\in\mathcal S_2}(c_1\cap c_2)$$ So you can just intersect individual circles and take union of the intersections.

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But the problem is for how to select individual circles from each group? I mean say that you selected c1 from S1 and then which circle from S2? Or do you have to check c1 with every circle in S2 and do same thing for all S1? How can you do that? – kotoll Aug 3 '12 at 21:43
@kamuran: you don't select them; you intersect every single pair. That's what the union means: you take union over all pairs. You can write it as $\bigcup_{(c_1,c_2)\in \mathcal S_1\times \mathcal S_2}$, if you prefer... – tomasz Aug 4 '12 at 1:06