第十二周项目4——圆，点的关系（两个交点的坐标）

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（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：

代码如下：

#include <iostream>#include <Cmath>using namespace std;class Circle;class Point{protected:    double x,y;public:    Point(int xx=0,int yy=0):x(xx),y(yy) {}    friend ostream& operator<<(ostream&output,const Point &c);    friend void Point_intersection(Point &p,Circle &c,Point &p1,Point &p2); //求交点的友元函数};ostream& operator<<(ostream &output,const Point &c){    output<<"("<<c.x<<","<<c.y<<")"<<endl;    return output;}class Circle:public Point{protected:    double r;public:    Circle(int xx=0,int yy=0,double rr=0):Point(xx,yy),r(rr){};    friend ostream& operator<<(ostream&output,const Circle &c);    friend void Point_intersection(Point &p,Circle &c,Point &p1,Point &p2); //求交点的友元函数};ostream& operator<<(ostream&output,const Circle &c){    output<<"("<<c.x<<","<<c.y<<","<<c.r<<")"<<endl;    return output;}void Point_intersection(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=((c.y-p.y)/(c.x-p.x))*(p1.x-p.x)+p.y;     p2.y=((c.y-p.y)/(c.x-p.x))*(p2.x-p.x)+p.y;}int main( ){    Circle c(3,2,4);    Point p(10,10),p1,p2;    Point_intersection(p,c,p1,p2);    cout<<"点p"<<p<<"与圆c"<<c<<"的圆心确定的一条直线与圆的两个交点："<<endl;    cout<<"交点1: "<<p1<<endl;    cout<<"交点2: "<<p2<<endl;    return 0;}

总结：

没有公式真心算不出来

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