不知道哪错了，请帮忙看看指错~~
#include<stdio.h>
#include<math.h>
#define PI 3.141592653589793
int main()
{
   double x1, y1, x2, y2, x3, y3, d, e, r;
   while(scanf("%lf%lf%lf%lf%lf%lf", &x1, &y1, &x2, &y2, &x3, &y3) == 6)
   {
       d = ((x2 * x2- x3 * x3) * (y2 - y1) - (x1 * x1 - x2 * x2) * (y3 - y2)) / ((x1 - x2) * (y3 - y2) - (x2 - x3) * (y2 - y1));
       e = (x1 * x1 - x2 * x2 + d * (x1 - x2)) / (y2 - y1);
       r = sqrt((x1 + d / 2) * (x1 + d / 2) + (y1 + e / 2) * (y1 + e / 2));
       printf("%.2lf\n", 2 * PI * r);
    }
    return 0;
}


/*(a, b)设为圆心的坐标，x1, y1, x2, y2, x3, y3分别为不共线三个点的坐标，已知（x - a）^ 2 + (y - b) ^ 2 = r ^ 2
拆开来就是x^2 - 2ax + a ^ 2 + y ^ 2 - 2by + b ^ 2 - r ^ 2 = 0,令d = -2a， e = -2b， f = a ^ 2 + b ^ 2 - r ^ 2
即为 x ^ 2 + dx + ey + f = 0,所以有式子 x1^ 2 + dx1 + ey1 + f = 0……1
                                         x2 ^ 2 + dx2 + ey2 + f = 0……2
                                          x3 ^ 2 + dx3 + ey3 + f = 0……3
联立1与2得， （x1 ^ 2 - x2 ^ 2） + d(x1 - x2) + e(y1 - y2) = 0…… 4
联立3与2得，（x2 ^ 2 - x3 ^ 2） + d(x2 - x3) + e(y2 - y3) = 0……5
4式即， e = ((x1 ^ 2 - x2 ^ 2) + d(x1 - x2)) / (y2 - y1)
5式即， e = ((x2 ^ 2 - x3 ^ 2) + d(x2 - x3)) / (y3 - y2)
4式-5式得，（x1 ^ 2 - x2 ^ 2） / (y2 - y1) - (x2 ^ 2 - x3 ^ 2) / (y3 - y2) + d((x1 - x2) / (y2 - y1) - (x2 - x3) / (y3 - y2)) = 0
即  d = ((x2 * x2- x3 * x3) * (y2 - y1) - (x1 * x1 - x2 * x2) * (y3 - y2)) / ((x1 - x2) * (y3 - y2) - (x2 - x3) * (y2 - y1))
   e = ((x1 * x1 - x2 * x2 + d * (x1 - x2)) / (y2 - y1) */