已知三点求圆心和半径-奇遇探秘-群侠问道-武侠网游活动门户

https://blog.csdn.net/youhuakongzhi/article/details/86474619

https://blog.csdn.net/qq_43572555/article/details/103470968

很简单的推导，以后用的时候就不用浪费时间了，

圆的一般方程为：

三个已知点为(x1,y1) (x2,y2) (x3,y3)

则圆心(x,y)和半径r的计算结果为：

需要注意，如果三个点共线，那么这三个点肯定无法形成圆，这一问题可通过上式的A来判定，A=0说明三点共线；

#include

using namespace std;

int main()

{

//已知三个点确定圆的半径和圆心

double x1,x2,x3,y1,y2,y3,x,y,r,A,B,C,D;

cin>>x1>>y1>>x2>>y2>>x3>>y3;

A=x1*(y2-y3)-y1*(x2-x3)+x2*y3-x3*y2;

B=(x1*x1+y1*y1)*(y3-y2)+(x2*x2+y2*y2)*(y1-y3)+(x3*x3+y3*y3)*(y2-y1);

C=(x1*x1+y1*y1)*(x2-x3)+(x2*x2+y2*y2)*(x3-x1)+(x3*x3+y3*y3)*(x1-x2);

D=(x1*x1+y1*y1)*(x3*y2-x2*y3)+(x2*x2+y2*y2)*(x1*y3-x3*y1)+(x3*x3+y3*y3)*(x2*y1-x1*y2);

x=-B/(2*A);

y=-C/(2*A);

r=sqrt((B*B+C*C-4*A*D)/(4*A*A));

//-1表示圆不存在

if(!A)

cout<<"-1"<

else

printf("%.4lf %.4lf %.4lf\n",x,y,r);

return 0;

}

为加快运算速度，避免重复计算，把上面代码中重复计算的部分，提取出来，对于实时图像计算，分秒必争：

void xxxxx()

{

float x1,x2,x3,y1,y2,y3,x,y,r,A,B,C,D;

float x1x1 = x1*x1;

float y1y1 = y1*y1;

float x2x2 = x2*x2;

float y2y2 = y2*y2;

float x3x3 = x3*x3;

float y3y3 = y3*y3;

float x2y3 = x2*y3;

float x3y2 = x3*y2;

float x2_x3 = x2-x3;

float y2_y3 = y2-y3;

float x1x1py1y1 = x1x1 + y1y1;

float x2x2py2y2 = x2x2 + y2y2;

float x3x3py3y3 = x3x3 + y3y3;

A = x1 * y2_y3 - y1 * x2_x3 + x2y3 - x3y2;

B = x1x1py1y1 * (-y2_y3) + x2x2py2y2 * (y1-y3) + x3x3py3y3 * (y2-y1);

C = x1x1py1y1 * x2_x3 + x2x2py2y2 * (x3 - x1) + x3x3py3y3 * (x1-x2);

D = x1x1py1y1 * (x3y2 - x2y3) + x2x2py2y2 * (x1*y3 - x3*y1) + x3x3py3y3 * (x2*y1-x1*y2);