第十二周 项目4-点，圆类的类（3）

/* 
 * Copyright (c) 2014, 烟台大学计算机学院 
 * All rights reserved. 
 * 文件名称：test.cpp 
 * 作    者：呼亚萍 
 * 完成日期：2015年5月27日 
 * 版 本 号：v1.0 
 * 
 * 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：相应的程序
 * 程序输出：对应得结果
 */

#include<iostream>
#include<cmath>
using namespace std;
class Circle;//在Point类中引用，所以提前注明
class Point
{
public:
    Point()
    {
        x=0;
        y=0;
    }
    Point(double a,double b):x(a),y(b) {}
    friend ostream& operator<<(ostream&,Point&);
    friend void crossover_point(Point&p,Circle&c,Point&p1,Point&p2);
protected:
    double x;
    double y;
};
ostream& operator<<(ostream&output,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&,Circle&);
    friend void crossover_point(Point&p,Circle&c,Point&p1,Point&p2);//求交点的友员函数
private:
    double radius;
};
ostream& operator<<(ostream&output,Circle&c)
{
    output<<"("<<c.x<<","<<c.y<<")";
    cout<<"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;

}

运算结果：

知识点总结：

与数学结合，类的派生，求点与圆心连线与圆的两个交点！

学习心得：

将其他学科与c++联系起来学习！