第12周 项目四-点、圆关系（6）

（6）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。
提示：

代码：

#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struct DoublePoint;  //也先声明，Point中声明友元函数crossover_point中要用到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 DoublePoint crossover_point(Point &p,Circle &c) ;  //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 DoublePoint crossover_point(Point &p,Circle &c) ;  //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;}struct DoublePoint   //专门用于返回值的结构体类型{    Point p1;    Point p2;};//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//方案2：结果返回到DoublePoint类型的结构体中//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DoublePoint crossover_point(Point &p, Circle &c){    DoublePoint pp;    pp.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)))));    pp.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)))));    pp.p1.y = (p.y + (pp.p1.x -p.x)*(c.y-p.y)/(c.x-p.x));    pp.p2.y = (p.y + (pp.p2.x -p.x)*(c.y-p.y)/(c.x-p.x));    return pp;}int main( ){    Circle c1(3,2,4);    Point p1(1,1);    DoublePoint pp;    pp = crossover_point(p1,c1);    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: "<<pp.p1<<endl;    cout<<"交点2: "<<pp.p2<<endl;    return 0;}

运行结果：

知识点总结：

学了类 不要忘了结构体。。。