C++ 已知两点坐标和半径求圆心坐标程序

墨蓝 2022-05-08 08:58 327阅读 0赞

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 <iostream>
    #include <fstream>
    #include <cmath>
    
    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);
        }
    }

原文：https://blog.csdn.net/liumoude6/article/details/78114255