Simplest Formula to Find Equation of A Circle from 3 Points Given

**samtcs** · September 24th 2009, 08:35 PM

I need a simplest and 100% correct formula (or method) to find equation of a circle with 3 points given.

**red_dog** · September 25th 2009, 01:22 AM

See this: http://home.att.net/~srschmitt/circle3pts.html

**HallsofIvy** · September 27th 2009, 06:23 PM

Originally Posted by samtcs

I need a simplest and 100% correct formula (or method) to find equation of a circle with 3 points given.

Given that $(x_1, y_1)$ , $(x_2,y_2)$ , and $(x_2,y_3)$ are points on the plane, then $((x_1+x_2)/2, (y_1+y_2)/2)$ and $((x_1+x_3)/2, (y_1+y_3)/2)$ are midpoints of chords in the circle. The two chords have slopes $(y_1- y_2)/(x_1- x_2)$ and $(y_1- y_3)/(x_1-x_3)$ respectively so that $-(x_1-x_2)/(y-1-y_2)$ and $-(x_1-x_3)/(y_1-y_3)$ are the slopes of the perpendicular lines.

That means that $y= -(x_1-x_2)/(y_1-y_2)(x- x_1)+ y_1$ and $y= -(x_1- x_3)/(y_1-y_2)(x-x_1)+ y_1$ are equations of the radii through the midpoints and so intersect the center, $(x_0,y_0)$ . Solve those two equations simultaneously to find $x= x_0$ and $y= y_0$ . Of course, once you have that, the radius of the circle is given by $\sqrt{(x_1-x_0)^2+ (y_1- y_0)^2}$ and then the equation of the circle is $(x- x_0)^2+ (y_1- y_0)^2= r^2$ .

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