题解 | 直线与圆交点间距

我认为这题恶心的地方在于他给出的直线AB中的AB是直线上的任意2点而不是交点，在计算过程中很容易在最后求弦长的时候将半径写为a到b的距离

思路：求出圆心到直线的距离后，利用勾股定理将半径的平方减去距离的平方即可.

#include <bits/stdc++.h>
using namespace std;

struct point{
    double x,y;
    point(double A,double B){
        x=A,y=B;
    }
    point() = default;
};

struct line{
    point point_A,point_B;
    line(point A,point B){
        point_A = A,point_B = B;
    }
    line() = default;
};

struct Circle{
    point O;
    int r;
    Circle(point A,int B){
        O=A,r=B;
    }
    Circle() = default;
};

double getDistance(const Circle& circle, const line& l) {
    double c1,c2,c3;
    c1=l.point_A.y-l.point_B.y;
    c2=-(l.point_A.x-l.point_B.x);
    c3=-(c2*l.point_A.y)-c1*l.point_A.x;
    double dis = fabs(c1*circle.O.x+c2*circle.O.y+c3)/sqrt(c1*c1+c2*c2);
    if(dis == 0){
        return 2*circle.r;
    }else if(dis == circle.r){
        return 0.0;
    }
    double len;
    len = sqrt(circle.r*circle.r - dis*dis);
    len = 2*len;
    return len;


    // 请在这里实现你的代码
}





















































































































































































int main() {
    double ox, oy, r;
    double x1, y1, x2, y2;
    
    cin >> ox >> oy >> r;
    cin >> x1 >> y1 >> x2 >> y2;
    
    point center(ox, oy);
    Circle circle(center, (int)r);
    
    point p1(x1, y1);
    point p2(x2, y2);
    line l(p1, p2);
    
    double result = getDistance(circle, l);
    cout << fixed << setprecision(6) << result << endl;
    
    return 0;
}