已知三点求圆心和半径
程序员文章站
2022-05-21 09:25:11
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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)
则圆心和半径为:
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 <iostream>
#include <math.h>
#include <iomanip>
#include <stdio.h>
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"<<endl;
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);
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"<<endl;
else
printf("%.4lf %.4lf %.4lf\n",x,y,r);
}