已知两点坐标和半径求圆心坐标程序C++

已知圆上的两点坐标和半径，求圆心。

数学分析：这个题目，涉及到简单的数学问题，但是计算比较繁琐。
假设已知圆上的两点坐标分别为N(X1，Y1)和M(X2，Y2)，半径为R，圆心坐标为o(a,b),根据数学知识可得到：
(x1-a)^2 + (y1-b)^2 = R^2----(1)式
(x2-a)^2 + (y2-b)^2 = R^2----(2)式
分别展开上述两个式子得到
(x1)^2 - 2*x1*a + a^2 + (y1)^2 - 2*y1*b + b^2 = R^2   ----(3)式
(x2)^2 - 2*x2*a + a^2 + (y2)^2 - 2*y2*b + b^2 = R^2   ----(4)式
(3)式 -  (4)式
得到：
 x1^2 - x2^2 + 2*(x2-x1)*a + y1^2 - y2^2 + 2*(y2-y1)*b = 0
变形得到：
a = (x2^2 - x1^2 + y2^2 - y1^2)/2/(x2-x1) - (y2-y1)/(x2-x2) * b

设：C1 = (x2^2 - x1^2 + y2^2 - y1^2)/2/(x2-x1)
设：C2 = (y2-y1)/(x2-x2)

a = c1 - c2 * b   ----(5)式
把(5)式代入(1)式，得到；
x1^2 - 2*x1*(C1-C2*b) + (C1-C2*b)^2 + y1^2 -2*y1*b + b^2 = R^2
展开简化为关于b的一元二次方程一般形式；
(C2^2+1)*b^2 + (2*x1*C1-2*C1*C2-2*y1)*b + x1^2-2*x1*C1+C1^2+y1^2-R^2 = 0

得到求b的方程组

二次项系数：A = (C2^2+1)
一次项系数：B = (2*x1*C1-2*C1*C2-2*y1)
常数项：    C = x1^2-2*x1*C1+C1^2+y1^2-R^2

已知两点坐标和半径求圆心坐标程序  
#include 
#include 
#include 

using namespace std;

ofstream fout;

typedef struct 
{
    double x;
    double y;
}Point;

double Y_Coordinates(double x,double y,double k,double x0);//4个参数
void Circle_Center(Point p1,Point p2,double dRadius);
bool Data_Validation(Point p1,Point p2,double dRadius);
void ReadData(Point &p1,Point &p2,double &dRadius);

void main()
{
    Point p1,p2;
    double dRadius = 0.0;

    ReadData(p1,p2,dRadius);

    fout.open("Result.txt");
    Circle_Center(p1,p2,dRadius);
    fout.close();
}

void Circle_Center(Point p1,Point p2,double dRadius)
{
    double k = 0.0,k_verticle = 0.0;
    double mid_x = 0.0,mid_y = 0.0;
    double a = 1.0;
    double b = 1.0;
    double c = 1.0;
    Point center1,center2;
    k = (p2.y - p1.y) / (p2.x - p1.x);
    if(k == 0)
    {
        center1.x = (p1.x + p2.x) / 2.0;
        center2.x = (p1.x + p2.x) / 2.0;
        center1.y = p1.y + sqrt(dRadius * dRadius -(p1.x - p2.x) * (p1.x - p2.x) / 4.0);
        center2.y = p2.y - sqrt(dRadius * dRadius -(p1.x - p2.x) * (p1.x - p2.x) / 4.0);
    }
    else
    {
        k_verticle = -1.0 / k;
        mid_x = (p1.x + p2.x) / 2.0;
        mid_y = (p1.y + p2.y) / 2.0;
        a = 1.0 + k_verticle * k_verticle;
        b = -2 * mid_x - k_verticle * k_verticle * (p1.x + p2.x);
        c = mid_x * mid_x + k_verticle * k_verticle * (p1.x + p2.x) * (p1.x + p2.x) / 4.0 - 
            (dRadius * dRadius - ((mid_x - p1.x) * (mid_x - p1.x) + (mid_y - p1.y) * (mid_y - p1.y)));
        
        center1.x = (-1.0 * b + sqrt(b * b -4 * a * c)) / (2 * a);
        center2.x = (-1.0 * b - sqrt(b * b -4 * a * c)) / (2 * a);
        center1.y = Y_Coordinates(mid_x,mid_y,k_verticle,center1.x);
        center2.y = Y_Coordinates(mid_x,mid_y,k_verticle,center2.x);
    }

    fout << center1.x << "    " << center1.y << endl;
    fout << center2.x << "    " << center2.y << endl;
}

double Y_Coordinates(double x,double y,double k,double x0)
{
    return k * x0 - k * x + y;
}

bool Data_Validation(Point p1,Point p2,double dRadius)
{
    double dDistance = 0.0;
    dDistance = sqrt((p1.x - p2.x) * (p1.x - p2.x) + (p1.y - p2.y) * (p1.y - p2.y));
    if(dDistance == 0.0)
    {
        cout << "\n输入了相同的点!\n";
        return false;
    }
    if((2 * dRadius) >= dDistance)
        return true;
    else
    {
        cout << "\n两点间距离大于直径!\n";
        return false;
    }
}

void ReadData(Point &p1,Point &p2,double &dRadius)
{
    cout << "请输入圆周上一点的坐标：";
    cin >> p1.x >> p1.y;
    cout << "请输入圆周上另一点的坐标：";
    cin >> p2.x >> p2.y;
    cout << "请输入圆的半径：";
    cin >> dRadius;

    if(! Data_Validation(p1,p2,dRadius))
    {
        cout << endl << "数据不合理!\n";
        exit(0);
    }
}