I would like to know how to solve this problem:

If you have overlapping circles $A$ and $B$ (with $B$'s area is at least half the area of $A$), what are the points of intersection of circle $A$ when the area of intersection between $A$ and $B$ is equal to half the area of $A$ (or any portion of $A$)?

  • $\begingroup$ That is a confusing problem. I wonder what we're supposed to do with "or any portion of $A$". What's so special about "half the area" if it can also be any portion? (Or half of any portion--the phrasing is ambiguous.) We can come up with relevant formulas, but if this question was posed by an instructor, perhaps they can clarify what it's really about. Did this have something to do with linear algebra, as the initial tag said? $\endgroup$ – David K Sep 11 at 11:16
  • $\begingroup$ Sorry, I don't have much of a math background so I was unsure how to phrase things. I just meant that "half" of circle A's area is arbitrary; the problem could be stated using any percentage of A's area. $\endgroup$ – Dano9700 Sep 11 at 20:57
  • $\begingroup$ So I suppose then this is a problem interesting to you for some purpose (not just for school or self-learning) and you had to try to come up with the wording yourself. That's important context for the question, and it helps people understand what you need. $\endgroup$ – David K Sep 12 at 0:34
  • $\begingroup$ Here is a question that I think is much the same problem, but formulated a bit differently, with another version of the answer: math.stackexchange.com/questions/2682600/… $\endgroup$ – David K Sep 12 at 0:37
  • $\begingroup$ Thanks. I was playing with two quarters and got to wondering... $\endgroup$ – Dano9700 Sep 12 at 1:28

The area of the intersection between your circles is given by: $$ A=r^2\cos^{-1}{d^2+r^2-R^2\over2dr} +R^2\cos^{-1}{d^2+R^2-r^2\over2dR} -{1\over2}\sqrt{(-d+r+R)(d+r-R)(d-r+R)(d+r+R)}, $$ where $R$ and $r$ are the radii and $d$ is the distance between the centres.

To find the intersections you need to plug the known value of $A$ into that formula and solve the resulting equation for $d$. But there is no analytical formula for that, you must find $d$ by some numerical method.

  • $\begingroup$ Thanks! I was hoping there was an analytical method, that's a bit disappointing. $\endgroup$ – Dano9700 Sep 11 at 21:02

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