点圆的关系---3

  输入代码：

/*
*Copyright (c)2015,烟台大学计算机与控制工程学院
*All rights reserved.
*文件名称：sum123.cpp
*作    者：林海云
*完成日期：2015年6月12日
*版 本 号：v2.0
*
*问题描述：与圆心相连的直线：给定一点p，其与圆心相连成的直线，会和圆有两个交点，如图。
            在上面定义的Point(点)类和Circle(圆)类基础上，设计一种方案，输出这两点的坐标。
*程序输入：无
*程序输出：按要求输出两圆的交点
*/
#include <iostream>
#include<Cmath>
using namespace std;
class Circle;
class Point
{
public:
    Point(double a=0,double b=0):x(a),y(b) {}
    friend ostream & operator<<(ostream &,const Point &);
    friend void crossover_point(Point &p,Circle &c, Point &p1,Point &p2 ) ;  //求交点的友元函数
protected:
    double x,y;
};
ostream & operator<<(ostream &output,const Point &p)
{
    output<<"["<<p.x<<","<<p.y<<"]";
    return output;
}

class Circle:public Point
{
public:
    Circle(double a=0,double b=0,double r=0):Point(a,b),radius(r) { }
    friend ostream &operator<<(ostream &,const Circle &);
    friend void crossover_point(Point &p,Circle &c, Point &p1,Point &p2 ) ;  //求交点的友元函数
protected:
    double radius;
};
//重载运算符“<<”，使之按规定的形式输出圆的信息
ostream &operator<<(ostream &output,const Circle &c)
{
    output<<"Center=["<<c.x<<", "<<c.y<<"], r="<<c.radius;
    return output;
}
void crossover_point(Point &p, Circle &c, Point &p1,Point &p2 )
{
    p1.x = (c.x + sqrt(c.radius*c.radius/(1+((c.y-p.y)/(c.x-p.x))*((c.y-p.y)/(c.x-p.x)))));
    p2.x = (c.x - sqrt(c.radius*c.radius/(1+((c.y-p.y)/(c.x-p.x))*((c.y-p.y)/(c.x-p.x)))));
    p1.y = (p.y + (p1.x -p.x)*(c.y-p.y)/(c.x-p.x));
    p2.y = (p.y + (p2.x -p.x)*(c.y-p.y)/(c.x-p.x));
}
int main( )
{
    Circle c1(3,2,4);
    Point p1(1,1),p2,p3;

    crossover_point(p1,c1, p2, p3);

    cout<<"点p1: "<<p1<<endl;
    cout<<"与圆c1: "<<c1<<endl;
    cout<<"的圆心相连，与圆交于两点，分别是："<<endl;
    cout<<"交点1: "<<p2<<endl;
    cout<<"交点2: "<<p3<<endl;
    return 0;
}

点圆的关系---3

原文：http://blog.csdn.net/linhaiyun_ytdx/article/details/46475507