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### Finding the Center of a Circle from Three Points

```
Date: 05/22/2000 at 10:18:50
From: Christian Furst
Subject: Center point of circle

I have the coordinates of three ordered points on a circle. I want to
find a way to define the circle's center. The purpose is to find a way
to draw the part of the circle that is connecting the three points (by
```

```
Date: 05/22/2000 at 15:34:50
From: Doctor Rob
Subject: Re: Center point of circle

Thanks for writing to Ask Dr. Math, Christian.

Let the equation of the circle be

(x-h)^2 + (y-k)^2 = r^2,

and substitute the three known points, getting 3 equations in 3
unknowns h, k, and r:

(x1-h)^2 + (y1-k)^2 = r^2
(x2-h)^2 + (y2-k)^2 = r^2
(x3-h)^2 + (y3-k)^2 = r^2

which you can solve simultaneously. First subtract the third equation
from the other two, thus eliminating r^2, h^2, and k^2. That will
leave you with 2 simultaneous linear equations in h and k to solve.
This you can do as long as the 3 points are not collinear. Then those
values of h and k can be used in the first equation to find the

r = sqrt[(x1-h)^2 + (y1-k)^2].

Example: Suppose a circle passes through the points (4,1), (-3,7), and
(5,-2). Then we know that:

(h-4)^2 + (k-1)^2 = r^2
(h+3)^2 + (k-7)^2 = r^2
(h-5)^2 + (k+2)^2 = r^2

Subtracting the first from the other two, you get:

(h+3)^2 - (h-4)^2 + (k-7)^2 - (k-1)^2 = 0,
(h-5)^2 - (h-4)^2 + (k+2)^2 - (k-1)^2 = 0,

h^2 + 6h + 9 - h^2 + 8h - 16 + k^2 - 14k + 49 - k^2 + 2k - 1 = 0
h^2 - 10h + 25 - h^2 + 8h - 16 + k^2 + 4k + 4 - k^2 + 2k - 1 = 0

14h - 12k + 41 = 0
-2h +  6k + 12 = 0

10h +  65 = 0
30h + 125 = 0

h = -13/2
k = -25/6

Then

r = sqrt[(4+13/2)^2 + (1+25/6)^2]
= sqrt[4930]/6

Thus the equation of the circle is:

(x+13/2)^2 + (y+25/6)^2 = 4930/36

Understood?

- Doctor Rob, The Math Forum
http://mathforum.org/dr.math/
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Associated Topics:
High School Basic Algebra
High School Conic Sections/Circles
High School Coordinate Plane Geometry
High School Geometry

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