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 knowing the center and radius).
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 radius: 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/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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