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

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