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Finding the Center of a Circle Given 3 Points


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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