What is the algorithm for finding the center of a circle from three points?

Question

I have three points on the circumference of a circle:

pt A = (A.x, A.y); pt B = (B.x, B.y); pt C = (C.x, C.y); 

How do I calculate the center of the circle?

Implementing it in Processing (Java).

I found the answer and implemented a working solution:

 pt circleCenter(pt A, pt B, pt C) {      float yDelta_a = B.y - A.y;     float xDelta_a = B.x - A.x;     float yDelta_b = C.y - B.y;     float xDelta_b = C.x - B.x;     pt center = P(0,0);      float aSlope = yDelta_a/xDelta_a;     float bSlope = yDelta_b/xDelta_b;       center.x = (aSlope*bSlope*(A.y - C.y) + bSlope*(A.x + B.x)         - aSlope*(B.x+C.x) )/(2* (bSlope-aSlope) );     center.y = -1*(center.x - (A.x+B.x)/2)/aSlope +  (A.y+B.y)/2;      return center;   } 

Answer 1

Here's my Java port, dodging the error condition when the determinant disappears with a very elegant IllegalArgumentException, my approach to coping with the "points are two far apart" or "points lie on a line" conditions. Also, this computes the radius (and copes with exceptional conditions) which your intersecting-slopes approach will not do.

public class CircleThree {    static final double TOL = 0.0000001;    public static Circle circleFromPoints(final Point p1, final Point p2, final Point p3)   {     final double offset = Math.pow(p2.x,2) + Math.pow(p2.y,2);     final double bc =   ( Math.pow(p1.x,2) + Math.pow(p1.y,2) - offset )/2.0;     final double cd =   (offset - Math.pow(p3.x, 2) - Math.pow(p3.y, 2))/2.0;     final double det =  (p1.x - p2.x) * (p2.y - p3.y) - (p2.x - p3.x)* (p1.y - p2.y);       if (Math.abs(det) < TOL) { throw new IllegalArgumentException("Yeah, lazy."); }      final double idet = 1/det;      final double centerx =  (bc * (p2.y - p3.y) - cd * (p1.y - p2.y)) * idet;     final double centery =  (cd * (p1.x - p2.x) - bc * (p2.x - p3.x)) * idet;     final double radius =         Math.sqrt( Math.pow(p2.x - centerx,2) + Math.pow(p2.y-centery,2));      return new Circle(new Point(centerx,centery),radius);   }    static class Circle   {     final Point center;     final double radius;     public Circle(Point center, double radius)     {       this.center = center; this.radius = radius;     }     @Override      public String toString()     {       return new StringBuilder().append("Center= ").append(center).append(", r=").append(radius).toString();     }   }    static class Point   {     final double x,y;      public Point(double x, double y)     {       this.x = x; this.y = y;     }     @Override     public String toString()     {       return "("+x+","+y+")";     }    }    public static void main(String[] args)   {     Point p1 = new Point(0.0,1.0);     Point p2 = new Point(1.0,0.0);     Point p3 = new Point(2.0,1.0);     Circle c = circleFromPoints(p1, p2, p3);     System.out.println(c);   }  } 

See algorithm from here:

void circle_vvv(circle *c) {     c->center.w = 1.0;     vertex *v1 = (vertex *)c->c.p1;     vertex *v2 = (vertex *)c->c.p2;     vertex *v3 = (vertex *)c->c.p3;     float bx = v1->xw; float by = v1->yw;     float cx = v2->xw; float cy = v2->yw;     float dx = v3->xw; float dy = v3->yw;     float temp = cx*cx+cy*cy;     float bc = (bx*bx + by*by - temp)/2.0;     float cd = (temp - dx*dx - dy*dy)/2.0;     float det = (bx-cx)*(cy-dy)-(cx-dx)*(by-cy);     if (fabs(det) < 1.0e-6) {         c->center.xw = c->center.yw = 1.0;         c->center.w = 0.0;         c->v1 = *v1;         c->v2 = *v2;         c->v3 = *v3;         return;         }     det = 1/det;     c->center.xw = (bc*(cy-dy)-cd*(by-cy))*det;     c->center.yw = ((bx-cx)*cd-(cx-dx)*bc)*det;     cx = c->center.xw; cy = c->center.yw;     c->radius = sqrt((cx-bx)*(cx-bx)+(cy-by)*(cy-by)); } 

Answer 2

It can be a rather in depth calculation. There is a simple step-by-step here: http://paulbourke.net/geometry/circlesphere/. Once you have the equation of the circle, you can simply put it in a form involving H and K. The point (h,k) will be the center.

(scroll down a little ways at the link to get to the equations)