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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 

Best regards,

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 

    (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  

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)
  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   
Associated Topics:
College Conic Sections/Circles
High School Conic Sections/Circles

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