/*
*Copyright (c) 2014,煙臺大學計算機學院
*All rights reserved.
*檔名稱:Annpion.cpp
*作者:王耀鵬
*完成日期:2015年5月21日
*版本號:v1.0
*
*問題描述:以Point為基類，派生出一個Circle(圓)類。
*輸入描述:無。
*輸出描述:點與圓心所成直線與圓的兩個交點。
*/
#include <iostream>
#include<cmath>
using namespace std;
class Circle;
class Point
{
protected:
    double x,y;
public:
    Point (){}
    Point(double a,double b):x(a),y(b){}
    friend ostream &operator<<(ostream & out,const Point &p);
    friend void doublePoint (Point &p,Circle &c,Point &p1,Point &p2);
};
ostream &operator<<(ostream & out,const Point &p)
{
    cout<<"("<<p.x<<","<<p.y<<")";
    return out;
}
class Circle:public Point
{
private:
    double r;
public:
    Circle(double a,double b,double c):Point(a,b),r(c){}
    friend ostream &operator<<(ostream & out,const Circle &c);
    friend void doublePoint (Point &p,Circle &c,Point &p1,Point &p2);
};
ostream &operator<<(ostream & out,const Circle &c)
{
    cout<<"圓心為："<<"("<<c.x<<","<<c.y<<")"<<"半徑為："<<c.r;
    return out;
}
void doublePoint (Point &p,Circle &c,Point &p1,Point &p2)
{
    p1.x = (c.x + sqrt(c.r*c.r/(1+((c.y-p.y)/(c.x-p.x))*((c.y-p.y)/(c.x-p.x)))));
    p2.x = (c.x - sqrt(c.r*c.r/(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;

    doublePoint(p1,c1, p2, p3);

    cout<<"點p1: "<<p1<<endl;
    cout<<"與圓c1: "<<c1<<endl;
    cout<<"的圓心相連，與圓交於兩點，分別是："<<endl;
    cout<<"交點1: "<<p2<<endl;
    cout<<"交點2: "<<p3<<endl;
    return 0;
}


 

執行結果：