問題：
已知圓上三個點坐標分別為（x1,y1）、（x2,y2）、（x3,y3）
求圓半徑R和圓心坐標（X,Y）
X,Y,R為未知數,
x1,y1,x2,y2,x3,y3為常數
則由圓公式：
(x1-X)²+(y1-Y)²=R² (1)式
(x2-X)²+(y2-Y)²=R² (2)式
(x3-X)²+(y3-Y)²=R² (3)式
(1)-(2),就是左邊減左邊,右邊減右邊,得到
x1²-2Xx1+X²+(y1²-2Yy1+Y²)-(x2²-2Xx2+X²)-(y2²-2Yy2+Y²)=R²-R²
整理得
x1²-x2²-2*x1*X+2*x2*X+y12-y22-2*y1*Y+2*y2*Y=0
(2)-(3)整理得:
x2²-x3²-2*x2*X+2*x3*X+y22-y32-2*y2*Y+2y3*Y=0
再整理上面兩式得 (2x2-2x1)X+(2y2-2y1)Y=x2²-x1²+y2²-y1² (2x3-2x2)X+(2y3-2y2)Y=x3²-x2²+y3²-y2²
令： a=2x3-2x2;b=2y3-2y2;c=x3²-x2²+y3²-y2² e = 2x2-2x1;f=2y2-2y1;g=x2²-x1²+y2²-y1²
於是有 eX+fY=g aX+bY=c
解得 X=(gb-cf)\(eb-af) Y=(ag-ce)\(af-be)
R=sqrt((X-x1)*(X-x1)+(Y-y1)*(Y-y1))則圓心坐標為（X，Y），半徑為R
1 程序實現： 2 void Calculate_cicular(Point px1, Point px2, Point px3) 3 { 4 int x1, y1, x2, y2, x3, y3; 5 int a, b, c, g, e, f; 6 7 x1 = px1.x;//定義點的坐標 8 y1 = px1.y; 9 x2 = px2.x; 10 y2 = px2.y; 11 x3 = px3.x; 12 y3 = px3.y; 13 14 e = 2 * (x2 - x1); 15 f = 2 * (y2 - y1); 16 g = x2*x2 - x1*x1 + y2*y2 - y1*y1; 17 18 a = 2 * (x3 - x2); 19 b = 2 * (y3 - y2); 20 c = x3*x3 - x2*x2 + y3*y3 - y2*y2; 21 22 X = (g*b - c*f) / (e*b - a*f); 23 Y = (a*g - c*e) / (a*f - b*e); 24 25 R = sqrt((X-x1)*(X-x1)+(Y-y1)*(Y-y1)); 26 27 }