Intersecting CirclesFrom: Martin Subject: intersection of two circles Date: 2 Apr 1995 09:30:50 -0400 Hello Kings of Mathematics, I read about your group in a journal for network computing. The magazine told that it's possible to get a solution of certain mathematical problems from you. I hope the problem is difficult enough for you! The problem is, I do not think the problem is too difficult but for me. So I have been happy to read about your box. What can I do? - I decided to tell you the problem: The Problem =========== What is the solution to determine the intersection of two given circles (which don't have the same center and do not only touch in one point)? The formula of both circles are: (y-y01)^2 + (x-x01)^2 = R1^2 (eq.1) (y-y02)^2 + (x-x02)^2 = R2^2 (eq.2) where the center of both is M1: (x01 | y01) M2: (x02 | y02) The two points of intersection are A: (x1 | y1) and B: (x2 | y2). My Idea for a solution: ==================== In the first shot, I could not solve the two equations in the second power. I blame myself, but I think it is very tricky. Then I looked to the geometric circumstances, to get one linear equation for substitution. I thought: __ ____ The line AB is devided by the rectangle line M1M2 in two halfs. The slope ____ of M1M2 is m = y02-y01/x02-x01 and well known. So I get a third, linear equation like y = - 1/m * x + b (eq.3). But there must be another, better idea to grip eq. 1/2! I never eliminate the square roots during my evaluations! Can you help me? I'm sorry, it is a blame for me, but I would be happy to read your hints to me. I am looking forward to your PM an say thank you! Regards Martin Dortmund, Germany Date: 2 Apr 1995 16:50:57 -0400 From: Dr. Ken Subject: Re: intersection of two circles Hello there! Well, you certainly shouldn't feel bad that you couldn't solve this problem. I don't think it's as easy as it looks. Consider the fact that you could change the centers of the circles in such a way that the circles would intersect in no points, then one point, then two points, then one point, and then no points again. So the formula for the points of intersection is going to be quite complex in order to account for that. I fed the equations into Mathematica, and it gave me this answer. For brevity of notation, I changed the equations to (x-a)^2 + (y-b)^2 == R (x-c)^2 + (y-d)^2 == r. These are the two points: {{x -> (a^4 + a^2*b^2 - 2*a^3*c - b^2*c^2 + 2*a*c^3 - c^4 - 2*a^2*b*d + 2*b*c^2*d + a^2*d^2 - c^2*d^2 + a^2*r - 2*a*c*r + c^2*r - a^2*R + 2*a*c*R - c^2*R + b*((a - c)^2* (-a^4 - 2*a^2*b^2 - b^4 + 4*a^3*c + 4*a*b^2*c - 6*a^2*c^2 - 2*b^2*c^2 + 4*a*c^3 - c^4 + 4*a^2*b*d + 4*b^3*d - 8*a*b*c*d + 4*b*c^2*d - 2*a^2*d^2 - 6*b^2*d^2 + 4*a*c*d^2 - 2*c^2*d^2 + 4*b*d^3 - d^4 + 2*a^2*r + 2*b^2*r - 4*a*c*r + 2*c^2*r - 4*b*d*r + 2*d^2*r - r^2 + 2*a^2*R + 2*b^2*R - 4*a*c*R + 2*c^2*R - 4*b*d*R + 2*d^2*R + 2*r*R - R^2))^(1/2) - d*((a - c)^2*(-a^4 - 2*a^2*b^2 - b^4 + 4*a^3*c + 4*a*b^2*c - 6*a^2*c^2 - 2*b^2*c^2 + 4*a*c^3 - c^4 + 4*a^2*b*d + 4*b^3*d - 8*a*b*c*d + 4*b*c^2*d - 2*a^2*d^2 - 6*b^2*d^2 + 4*a*c*d^2 - 2*c^2*d^2 + 4*b*d^3 - d^4 + 2*a^2*r + 2*b^2*r - 4*a*c*r + 2*c^2*r - 4*b*d*r + 2*d^2*r - r^2 + 2*a^2*R + 2*b^2*R - 4*a*c*R + 2*c^2*R - 4*b*d*R + 2*d^2*R + 2*r*R - R^2 ))^(1/2))/(2*(a - c)*(a^2 + b^2 - 2*a*c + c^2 - 2*b*d + d^2)), y -> (a^2*b + b^3 - 2*a*b*c + b*c^2 + a^2*d - b^2*d - 2*a*c*d + c^2*d - b*d^2 + d^3 + b*r - d*r - b*R + d*R - ((a - c)^2*(-a^4 - 2*a^2*b^2 - b^4 + 4*a^3*c + 4*a*b^2*c - 6*a^2*c^2 - 2*b^2*c^2 + 4*a*c^3 - c^4 + 4*a^2*b*d + 4*b^3*d - 8*a*b*c*d + 4*b*c^2*d - 2*a^2*d^2 - 6*b^2*d^2 + 4*a*c*d^2 - 2*c^2*d^2 + 4*b*d^3 - d^4 + 2*a^2*r + 2*b^2*r - 4*a*c*r + 2*c^2*r - 4*b*d*r + 2*d^2*r - r^2 + 2*a^2*R + 2*b^2*R - 4*a*c*R + 2*c^2*R - 4*b*d*R + 2*d^2*R + 2*r*R - R^2))^(1/2))/ (2*(a^2 + b^2 - 2*a*c + c^2 - 2*b*d + d^2))}, {x -> (a^4 + a^2*b^2 - 2*a^3*c - b^2*c^2 + 2*a*c^3 - c^4 - 2*a^2*b*d + 2*b*c^2*d + a^2*d^2 - c^2*d^2 + a^2*r - 2*a*c*r + c^2*r - a^2*R + 2*a*c*R - c^2*R - b*((a - c)^2* (-a^4 - 2*a^2*b^2 - b^4 + 4*a^3*c + 4*a*b^2*c - 6*a^2*c^2 - 2*b^2*c^2 + 4*a*c^3 - c^4 + 4*a^2*b*d + 4*b^3*d - 8*a*b*c*d + 4*b*c^2*d - 2*a^2*d^2 - 6*b^2*d^2 + 4*a*c*d^2 - 2*c^2*d^2 + 4*b*d^3 - d^4 + 2*a^2*r + 2*b^2*r - 4*a*c*r + 2*c^2*r - 4*b*d*r + 2*d^2*r - r^2 + 2*a^2*R + 2*b^2*R - 4*a*c*R + 2*c^2*R - 4*b*d*R + 2*d^2*R + 2*r*R - R^2))^(1/2) + d*((a - c)^2*(-a^4 - 2*a^2*b^2 - b^4 + 4*a^3*c + 4*a*b^2*c - 6*a^2*c^2 - 2*b^2*c^2 + 4*a*c^3 - c^4 + 4*a^2*b*d + 4*b^3*d - 8*a*b*c*d + 4*b*c^2*d - 2*a^2*d^2 - 6*b^2*d^2 + 4*a*c*d^2 - 2*c^2*d^2 + 4*b*d^3 - d^4 + 2*a^2*r + 2*b^2*r - 4*a*c*r + 2*c^2*r - 4*b*d*r + 2*d^2*r - r^2 + 2*a^2*R + 2*b^2*R - 4*a*c*R + 2*c^2*R - 4*b*d*R + 2*d^2*R + 2*r*R - R^2))^(1/2))/ (2*(a - c)*(a^2 + b^2 - 2*a*c + c^2 - 2*b*d + d^2)), y -> (a^2*b + b^3 - 2*a*b*c + b*c^2 + a^2*d - b^2*d - 2*a*c*d + c^2*d - b*d^2 + d^3 + b*r - d*r - b*R + d*R + ((a - c)^2*(-a^4 - 2*a^2*b^2 - b^4 + 4*a^3*c + 4*a*b^2*c - 6*a^2*c^2 - 2*b^2*c^2 + 4*a*c^3 - c^4 + 4*a^2*b*d + 4*b^3*d - 8*a*b*c*d + 4*b*c^2*d - 2*a^2*d^2 - 6*b^2*d^2 + 4*a*c*d^2 - 2*c^2*d^2 + 4*b*d^3 - d^4 + 2*a^2*r + 2*b^2*r - 4*a*c*r + 2*c^2*r - 4*b*d*r + 2*d^2*r - r^2 + 2*a^2*R + 2*b^2*R - 4*a*c*R + 2*c^2*R - 4*b*d*R + 2*d^2*R + 2*r*R - R^2))^(1/2))/ (2*(a^2 + b^2 - 2*a*c + c^2 - 2*b*d + d^2))}} The computer couldn't simplify this any further. I'm not sure whether this will be useful to you, but I thought I'd give it to you anyway. Let us know if you don't understand the syntax of the formula. Thanks! -Ken "Dr." Math Date: 04/24/2003 at 18:33:29 From: Cesar Subject: Intersecting Circles (again) Dear Dr Math: I tried feeding these equations into Visual Basic and then verified the results using AutoCAD. I get close values but they're not accurate enough (e.g., 4.6356 instead of 4.9243). Is there another way to solve this problem? Date: 04/25/2003 at 10:45:08 From: Doctor George Subject: Re: Intersecting Circles (again) Hi Cesar, I would recommend a vector approach to finding the intersection points of two circles. First, let C1 and C2 be the centers of the circles with radii r1 and r2, and let d be the distance between C1 and C2. Now let V1 be the unit vector from C1 to C2, and let V2 be the unit vector perpendicular to V1. Also let V3 be the vector from C1 to one of the intersection points. Finally, let A be the angle between V1 and V3. From the law of cosines we know that r2^2 = r1^2 + d^2 - 2*r1*d*cos(A) With this equation we can solve for 'A'. The intersection points will be C1 + [r1*cos(A)]*V1 + [r1*sin(A)]*V2 C1 + [r1*cos(A)]*V1 - [r1*sin(A)]*V2 Does this make sense to you? Writing your program this way should make it much easier to debug. A separate thread in our archive, Intersections of Circles http://mathforum.org/library/drmath/view/51836.html shows an equivalent solution expressed algebraically, instead of with vectors. -Doctor George, The Math Forum http://mathforum.org/dr.math/ |
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