Intersection of Circles
Date: 01/16/2002 at 06:36:21 From: Peter Knoben Subject: Intersection of circles I have two circles, one with radius R and centerpoint (a,b) and one with radius r and centerpoint (c,d). These two circles intersect. How can I find the coordinates of the intersection point(s)? I already tried the circle formula: (x-a)^2+(y-b)^2 = R^2 and (x-c)^2+(y-d)^2 = r^2 but if I make an equation of these two formulas I get an equation with square roots and exponentials that I can not solve. If it's possible, I would prefer a solution without the use of sine, cosine, or tangent calculations. Best regards, Peter
Date: 01/16/2002 at 12:43:18 From: Doctor Peterson Subject: Re: Intersection of circles Hi, Peter. You can solve the equations you gave, if you approach it the right way: (x-a)^2 + (y-b)^2 = R^2 (x-c)^2 + (y-d)^2 = r^2 If you expand each of these and then subtract one from the other, you will eliminate the squares, and will be left with a linear equation that you can easily solve for y. Replace y in one of the original equations with that expression, and you have a quadratic you can solve for x. It will be extremely ugly, but is not really hard. When you are finished, compare your result to this, which I found by searching our archives for the words "intersection circles": Intersecting Circles http://mathforum.org/dr.math/problems/circintersect.html I found a clearer solution than Dr. Ken's computer-generated solution linked from that page. Here it is: Let the centers be: (a,b), (c,d) Let the radii be: r, s e = c - a [difference in x coordinates] f = d - b [difference in y coordinates] p = sqrt(e^2 + f^2) [distance between centers] k = (p^2 + r^2 - s^2)/(2p) [distance from center 1 to line joining points of intersection] x = a + ek/p + (f/p)sqrt(r^2 - k^2) y = b + fk/p - (e/p)sqrt(r^2 - k^2) OR x = a + ek/p - (f/p)sqrt(r^2 - k^2) y = b + fk/p + (e/p)sqrt(r^2 - k^2) I found this solution using translation and rotation to simplify the math. To do the rotation, I used the fact that sin(angle) = f/p cos(angle) = e/p - Doctor Peterson, The Math Forum http://mathforum.org/dr.math/
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