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Given 2 circles on a plane, how do you calculate the intersecting points?

In this example I can do the calculation using the equilateral triangles that are described by the intersection and centres of the 2 circles, however, I need a more general formula that will provide the coordinates of the 2 intersection points C and D, on circles that are not placed so conveniently, or of the same radius.

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3 Answers 3

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Solve the the given 2 equations of the circle. For more you can see these links:

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thanks, I can read the C example in there, but the math notation makes me glaze over. –  Slomojo May 17 '11 at 5:44
    
Do you know of any solutions for ellipses, instead of regular circles? –  999 Jan 27 '12 at 15:41
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Each circle can be described by an equation of the form $(x-h)^2+(y-k)^2=r^2$, where $(h,k)$ is the center of the circle and $r$ is its radius. Given the equations of the two circles, expanding the squared terms and subtracting the equations should yield an equation of the line through the two points of intersection. Solving that for $x$ or $y$ and substituting into one of the original circle equations should yield the coordinates of your two points of intersection.

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That line is called the radical line. –  J. M. May 17 '11 at 5:24
    
@J.M.: Did you read my mind to know that I was sitting here trying to remember what it was called? Notably, subtracting one circle equation from the other always gives an equation for the radical line, which exists whether or not the circles intersect. –  Isaac May 17 '11 at 5:24
    
the idea of solving circle-related problems with the radical line has certainly popped up a number of times in math.SE ... :D –  J. M. May 17 '11 at 6:05
    
@Isaac - I tried doing what you suggested, but after solving the radical line for x and substituting into one of the circle equations, I get a pretty gigantic mess that has to be solved for y (namely with circles (x-i)^2 + (y-j)^2 = r_1^2 and (x-h)^2 + (y-k)^2 = r_2^2, I get ((r_1^2 - r_2^2 - i^2 + h^2 - j^2 + k^2 - y*(-2j + 2k))/(-2i + h) - i)^2 + (y-j)^2 = r_1^2. Solving this for y doesn't look like any fun at all... (Sorry for the nasty formatting, I don't think tex is allowed in comments?). Is there some kind of simplification I am missing to allow this to be more sane? –  David Doria Feb 13 '13 at 21:13
    
@DavidDoria: $\TeX$ should work in comments, between dollar signs. In the fully-general case, I would expect things to be fairly ugly; with specific numbers, it's much more manageable. –  Isaac Feb 15 '13 at 7:02
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@Isaac: You can check whether circles are intersecting by comparing distance between circle centers to the summ of their radiuses. E.g.

$$\max(R_1,R_2) - \min(R_1,R_2) \le AB \le R_1 + R_2 \Rightarrow \text{ circles are intersecting}$$

$$AB = 0 \text{ and } R_1 = R_2 \Rightarrow \text{circles are the same}$$

In other cases no intersection.

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