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

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

很簡單的推導(dǎo)，以后用的時候就不用浪費(fèi)時間了，

圓的一般方程為：

三個已知點(diǎn)為(x1,y1)? ?(x2,y2)? (x3,y3)

則圓心(x,y)和半徑r的計算結(jié)果為：

需要注意，如果三個點(diǎn)共線，那么這三個點(diǎn)肯定無法形成圓，這一問題可通過上式的A來判定，A=0說明三點(diǎn)共線；

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#include <iostream> #include <math.h> #include <iomanip> #include <stdio.h> using namespace std; int main() {//已知三個點(diǎn)確定圓的半徑和圓心 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表示圓不存在 if(!A)cout<<"-1"<<endl;elseprintf("%.4lf %.4lf %.4lf\n",x,y,r);return 0; }

為加快運(yùn)算速度，避免重復(fù)計算，把上面代碼中重復(fù)計算的部分，提取出來，對于實(shí)時圖像計算，分秒必爭：

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表示圓不存在if(!A)cout<<"-1"<<endl;elseprintf("%.4lf %.4lf %.4lf\n",x,y,r);}

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