帮我用C++实现最小二乘法圆拟合，并给出实际代码

好的，以下是最小二乘法圆拟合的 C 代码：

#include <stdio.h>
#include <math.h>

#define N 10 // 数据点个数

int main()
{
// 定义数据点坐标
double x[N] = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10};
double y[N] = {2, 4, 5, 4, 6, 8, 9, 8, 10, 12};

// 计算数据点坐标平均值
double x_avg = , y_avg = ;
for (int i = ; i < N; i++) {
x_avg += x[i];
y_avg += y[i];
}
x_avg /= N;
y_avg /= N;

// 计算各项系数
double S_xx = , S_yy = , S_xy = , S_xxx = , S_yyy = , S_xxy = , S_xyy = ;
for (int i = ; i < N; i++) {
double xi = x[i] - x_avg;
double yi = y[i] - y_avg;
S_xx += xi * xi;
S_yy += yi * yi;
S_xy += xi * yi;
S_xxx += xi * xi * xi;
S_yyy += yi * yi * yi;
S_xxy += xi * xi * yi;
S_xyy += xi * yi * yi;
}

double C = (S_xxx + S_xyy) / (2 * S_xy);
double D = (S_yyy + S_xxy) / (2 * S_xy);
double E = S_xx + S_yy;
double F = -2 * S_xy;

// 计算圆心坐标和半径
double a = (D * F - C * E) / (F * F - 4 * C * E);
double b = (D * F + C * E) / (F * F - 4 * C * E);
double r = sqrt((S_xx + S_yy) / N + (C * C + D * D - E) / (N * (F * F - 4 * C * E)));

// 输出结果
printf("圆心坐标：(%lf, %lf)\n", a + x_avg, b + y_avg);
printf("半径：%.2lf\n", r);

return ;
}

希望能对你有所帮助！