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I've been wondering if the following has a generalisation, and if so, what is it? I couldn't arrive at a result, partly because the calculation was too much to do and I didn't know in what direction to proceed.

Consider two circles, $ (x-a)² + (y-b)² = c² $ and $ (x-p)² + (y-q)² = r² $. If they intersect, what is the area of their common region? (or, area between the two circles?)

One condition that can be imposed is that the distance between their centres is definitely less than the sum of their radii, only then will they intersect.

How do I proceed from here? Please point me in the right direction, and a detailed solution and explanation would help too. Thanks.

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The area of overlap is the union of two circular segments. $h$ in the formula is half $r+c$ minus the distance between the centers. You can get Wikipedia's $c$ by the law of cosines, then the central angles, then the area.

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  • $\begingroup$ I'm rather interested in the final result. What is it? I was wondering if it's something in the form of a determinant, or worth reaching at. $\endgroup$ – arya_stark May 2 '18 at 17:04
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First, find points of intersection, which will be the solutions of this system: $$\begin{cases} (x-a)^2+(y-b)^2=c^2 \\ -2ax-2by+2px+2qy=c^2-r^2-a^2-b^2+p^2+q^2 \\ \end{cases} $$ Once you found them, find the length of chord that connects intersection points. Let's say the chord length is $l$. The common area will be two circular segments. The area of each segment can be calculated using radius and chord length. See, for example, this: https://planetcalc.com/1421/

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  • $\begingroup$ I'm rather interested in the final result. What is it? I was wondering if it's something in the form of a determinant, or worth reaching at. $\endgroup$ – arya_stark May 2 '18 at 17:03

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