点圆的关系---3

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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;}
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