POJ 1329 Circle Through Three Points

題意：給你三個圓上的點，求圓的兩種表達式。首先普及知識：

外接圓半徑:公式:a/sinA=b/sinB=c/sinC=2R (R就是外接圓半徑)
本題可以這樣：①．先利用余弦定理：a^2=b^2+c^2－2bc·cosA
求出：cosA=(b^2+c^2－a^2)/2bc 在利用公式：sinA^2+cosA^2=1
確定 sinA=根號(1－cosA^2) =根號[(a^2+b^2+c^2)^2-2(a^4+b^4+c^4)]/(2bc)
然後代入 a/sinA=2R求出R． R=abc/根號[(a^2+b^2+c^2)^2-2(a^4+b^4+c^4)]       
 
定義：設平面上的三點A(x1,y1),B(x2,y2),C(x3,y3),定義
S(A,B,C) = (x1-x3)*(y2-y3) - (y1-y3)*(x2-x3)

已知三角形的三個頂點為A(x1,y1),B(x2,y2),C(x3,y3)，則該三角形的外心為：
           S((x1*x1+y1*y1, y1), (x2*x2+y2*y2, y2), (x3*x3+y3*y3, y3))
x0 = -----------------------------------------------------------
                       2*S(A,B,C)

         S((x1,x1*x1+y1*y1), (x2, x2*x2+y2*y2), (x3, x3*x3+y3*y3))
y0 = -----------------------------------------------------------
            2*S(A,B,C)

把圓心的坐標和半徑求出來之後就輸出。

代碼：

[cpp] 
#include<iostream> 
#include<cmath> 
using namespace std; 
double Dist(double x,double y,double x1,double y1) 
{ 
       return (x-x1)*(x-x1)+(y-y1)*(y-y1); 
} 
double Sn(double x1,double y1,double x2,double y2,double x3,double y3) 
{ 
       return (x1-x3)*(y2-y3)-(y1-y3)*(x2-x3); 
} 
void Fn(double x) 
{ 
     if(x<0) printf(" + %.3lf",-x); 
     else printf(" - %.3lf",x); 
} 
int main() 
{ 
    double x1,x2,x3,y1,y2,y3,x,y; 
    while( scanf("%lf%lf%lf%lf%lf%lf",&x1,&y1,&x2,&y2,&x3,&y3)!=EOF){ 
           double a=Dist(x1,y1,x2,y2); 
           double b=Dist(x2,y2,x3,y3); 
           double c=Dist(x3,y3,x1,y1); 
           double r=sqrt(a*b*c)/sqrt((a+b+c)*(a+b+c)-2*(a*a+b*b+c*c)); 
         //  printf("%.lf\n",r); 
           double s=Sn(x1,y1,x2,y2,x3,y3); 
           double s1=Sn( x1*x1+y1*y1, y1, x2*x2+y2*y2, y2, x3*x3+y3*y3, y3); 
           double s2=Sn( x1,x1*x1+y1*y1, x2, x2*x2+y2*y2, x3, x3*x3+y3*y3); 
           x=s1/(2*s); www.2cto.com
           y=s2/(2*s); 
         
           printf("(x"); 
           Fn(x); 
           printf(")^2 + (y"); 
           Fn(y); 
           printf(")^2 = %.3lf^2\n",r); 
           printf("x^2 + y^2"); 
           Fn(2*x); printf("x"); 
           Fn(2*y); printf("y"); 
           Fn(r*r-x*x-y*y); 
           printf(" = 0\n\n"); 
    } 
    return 0; 
}