思路：

http://blog.csdn.net/commonc/article/details/52291822

（照着算法步骤写……）

已知三点共圆 求圆心的时候 就设一下圆心坐标(x,y) 解个方程就好了

//By SiriusRen

#include <cmath>

#include <cstdio>

#include <algorithm>

using namespace std;

int n;double R,tempx,tempy,tempz,tmpx,tmpy,tmpz;

struct Point{double x,y;}point[100050],Ans;

double Sqr(double x){return x*x;}

double dis(Point a,Point b){return sqrt(Sqr(a.x-b.x)+Sqr(a.y-b.y));}

bool in_circle(Point x){return dis(Ans,x)<=R;}

int main(){

    scanf("%d",&n);

    for(int i=1;i<=n;i++)scanf("%lf%lf",&point[i].x,&point[i].y);

    random_shuffle(point+1,point+n);

    for(int i=1;i<=n;i++)if(!in_circle(point[i])){

        Ans.x=point[i].x,Ans.y=point[i].y,R=0;

        for(int j=1;j<i;j++)if(!in_circle(point[j])){

            Ans.x=(point[i].x+point[j].x)/2;

            Ans.y=(point[i].y+point[j].y)/2;

            R=dis(Ans,point[j]);

            for(int k=1;k<j;k++)if(!in_circle(point[k])){

                tempz=point[j].x-point[i].x;

                tempx=2*(point[i].y-point[j].y)/tempz;

                tempy=(Sqr(point[j].x)+Sqr(point[j].y)-Sqr(point[i].x)-Sqr(point[i].y))/tempz;

                tmpz=point[k].x-point[j].x;

                tmpx=2*(point[j].y-point[k].y)/tmpz;

                tmpy=(Sqr(point[k].x)+Sqr(point[k].y)-Sqr(point[j].x)-Sqr(point[j].y))/tmpz;

                Ans.y=(tmpy-tempy)/(tempx-tmpx);

                Ans.x=(tempx*Ans.y+tempy)/2;

                R=dis(Ans,point[j]);

            }

        }

    }

    printf("%f\n%f %f\n",R,Ans.x,Ans.y);

}