第十三周项目3--与圆心相连的直线

/*
Copyright(c)烟台大学计算机与控制工程学院
作者:刘慧艳
完成日期：22014.05.25
版本号：V1.1
问题描述：在项目1中定义的Point(点)类和Circle(圆)类基础上，
          设计一种方案，输出给定一点p与圆心相连成的直线与圆的两个交点。
*/
#include <iostream>
#include<Cmath>
using namespace std;
class Circle;  //由于在Point中声明友元函数crossover_point中参数中用了Circle，需要提前声明
class Point
{
public:
    Point(double a=0,double b=0):x(a),y(b) {}		//构造函数
    friend ostream & operator<<(ostream &,const Point &);//重载运算符“<<”
    friend void crossover_point(Point &p,Circle &c, Point &p1,Point &p2 ) ;  //求交点的友元函数
protected:										 //受保护成员
    double x,y;
};
ostream & operator<<(ostream &output,const Point &p)
{
    output<<"["<<p.x<<","<<p.y<<"]";
    return output;
}

class Circle:public Point //circle是Point类的公用派生类
{
public:
    Circle(double a=0,double b=0,double r=0):Point(a,b),radius(r) { } //构造函数
    friend ostream &operator<<(ostream &,const Circle &);//重载运算符“<<”
    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;
}

//给定一点p，求出该点与圆c的圆心相连成的直线与圆的两个交点p1和p2
//方案1：利用引用类型的形式参数，注意，下面的p1和p2将“带回”求得的结果
//crossover_point函数已经声明为Point和Circle类的友元函数，类中私有成员可以直接访问
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;
}