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### Finding the Center of a Circle Given 3 Points

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Date: 05/25/2000 at 00:14:35
From: Alison Jaworski
Subject: finding the coordinates of the center of a circle

Hi,

Can you help me? If I have the x and y coordinates of 3 points - i.e.
(x1,y1), (x2,y2) and (x3,y3) - how do I find the coordinates of the
center of a circle on whose circumference the points lie?

Thank you.
```

```
Date: 05/25/2000 at 10:45:58
From: Doctor Rob
Subject: Re: finding the coordinates of the center of a circle

Thanks for writing to Ask Dr. Math, Alison.

Let (h,k) be the coordinates of the center of the circle, and r its
radius. Then the equation of the circle is:

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

Since the three points all lie on the circle, their coordinates will
satisfy this equation. That gives you three equations:

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

in the three unknowns h, k, and r. To solve these, subtract the first
from the other two. That will eliminate r, h^2, and k^2 from the last
two equations, leaving you with two simultaneous linear equations in
the two unknowns h and k. Solve these, and you'll have the coordinates
(h,k) of the center of the circle. Finally, set:

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

and you'll have everything you need to know about the circle.

This can all be done symbolically, of course, but you'll get some
pretty complicated expressions for h and k. The simplest forms of
these involve determinants, if you know what they are:

|x1^2+y1^2  y1  1|        |x1  x1^2+y1^2  1|
|x2^2+y2^2  y2  1|        |x2  x2^2+y2^2  1|
|x3^2+y3^2  y3  1|        |x3  x3^2+y3^2  1|
h = ------------------,   k = ------------------
|x1  y1  1|               |x1  y1  1|
2*|x2  y2  1|             2*|x2  y2  1|
|x3  y3  1|               |x3  y3  1|

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

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

10*h + 65 = 0
30*k + 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

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

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