第12周 【项目4 - 点、圆的关系】（3）

问题描述：

（1）先建立一个Point(点)类，包含数据成员x,y(坐标点)；

int main( )
{
	Circle c1(3,2,4),c2(4,5,5);      //c2应该大于c1
	Point p1(1,1),p2(3,-2),p3(7,3);  //分别位于c1内、上、外

	cout<<"圆c1: "<<c1;

	cout<<"点p1: "<<p1;
	cout<<"点p1在圆c1之"<<((locate(p1, c1)>0)?"外":((locate(p1, c1)<0)?"内":"上"))<<endl;

	cout<<"点p2: "<<p2;
	cout<<"点p2在圆c1之"<<((locate(p2, c1)>0)?"外":((locate(p2, c1)<0)?"内":"上"))<<endl;

	cout<<"点p3: "<<p3;
	cout<<"点p3在圆c1之"<<((locate(p3, c1)>0)?"外":((locate(p3, c1)<0)?"内":"上"))<<endl;
	return 0;
}

（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;
class Point{
protected:
    double x,y;
public:
    Point(){x=0,y=0;}
    Point(double a,double b):x(a),y(b){}
    double getx(){return x;}
    double gety(){return y;}
    double distance(const Point& c){
        double dx = x-c.x;
        double dy = y-c.y;
        return sqrt(dx*dx+dy*dy);
    }
    friend istream&operator>>(istream& in,Point & c);
    friend ostream&operator<<(ostream& out,const Point& c);
    friend void fun(Circle &c, Point &p1);
};
istream& operator>>(istream& in,Point & c){
    cout<<"请输入点坐标（x，y）格式\n";
    char a,b,e;
    do{
        cin>>a>>c.x>>b>>c.y>>e;
        if(a=='('&&b==','&&e==')')
            break;
        cout<<"格式错误请重新输入！\n";
    } while(1);
    return in;
}
ostream& operator<<(ostream& out,const Point & c){
    cout<<"("<<c.x<<","<<c.y<<")\n";
    return out;
}
class Circle:public Point{
private:
    double r;
public:
    Circle():r(0){Point();}
    Circle(double a,double b,double c):Point(a,b),r(c){}
    friend istream&operator>>(istream& in,Circle & c);
    friend ostream&operator<<(ostream& out,const Circle& c);
    friend int locate(Point&,Circle&  );
    bool operator>(const Circle&);
    bool operator<(const Circle&);
    bool operator>=(const Circle&);
    bool operator<=(const Circle&);
    bool operator==(const Circle&);
    bool operator!=(const Circle&);
    friend void fun(Circle &c, Point &p1);
    double area(){
        return 3.14159*r*r;
    }
};
void fun(Circle &c, Point &p){
    Point p1,p2;
    p1.x = (c.x + sqrt(c.r*c.r/(1+((c.y-p.gety())/(c.x-p.getx()))*((c.y-p.gety())/(c.x-p.getx())))));
    p2.x = (c.x - sqrt(c.r*c.r/(1+((c.y-p.gety())/(c.x-p.getx()))*((c.y-p.gety())/(c.x-p.getx())))));
    p1.y = (p.gety() + (p1.x -p.getx())*(c.y-p.gety())/(c.x-p.getx()));
    p2.y = (p.gety() + (p2.x -p.gety())*(c.y-p.gety())/(c.x-p.getx()));
    cout<<"点p"<<p<<"与圆"<<c<<"相交于"<<p1<<"和"<<p2<<"两点\n";
}
bool Circle::operator>(const Circle& c){
    if(r>c.r)
        return true;
    return false;
}
bool Circle::operator<(const Circle& c){
    if(r<c.r)
        return true;
    return false;
}
bool Circle::operator>=(const Circle& c){
    if(r<c.r)
        return false;
    return true;
}
bool Circle::operator<=(const Circle& c){
    if(r>c.r)
        return false;
    return true;
}
bool Circle::operator==(const Circle& c){
    if(r==c.r)
        return true;
    return false;
}
bool Circle::operator!=(const Circle& c){
    if(r==c.r)
        return false;
    return true;
}
istream&operator>>(istream& in,Circle& c){
    Point d;
    cin>>d;
    c.x=d.getx();
    c.y=d.gety();
    cout<<"请输入半径：";
    cin>>c.r;
    return in;
}
ostream&operator<<(ostream& out,const Circle& c){
    cout<<"("<<c.x<<","<<c.y<<") 半径="<<c.r<<'\12';
    return out;
}
int locate(Point&t2,Circle&t1){
    Point t=t1;
    double x=t1.distance(t2);
    if (abs(x-t1.r)<1e-7)
        return 0;
    else if (x<t1.r)
        return -1;
    else
        return 1;
}
int main( ){
	Circle c1(3,2,4);
	Point p;
	cin>>p;
	fun(c1,p);
	return 0;
}

运行结果：