Finding the Center of a Circle Given 3 PointsDate: 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/ |
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