C++ 实现求两圆外公切线

下面是一个简单的 C++ 实现，可以求解两个圆的外公切线。其中，Circle 类表示圆，包含圆心坐标和半径信息，Line 类表示直线，包含直线上的两个点的坐标信息。

#include <iostream>
#include <cmath>

using namespace std;

const double eps = 1e-8;

struct Point {
    double x, y;
    Point() {}
    Point(double x, double y) : x(x), y(y) {}
};

struct Circle {
    Point c;
    double r;
    Circle() {}
    Circle(Point c, double r) : c(c), r(r) {}
};

struct Line {
    Point p1, p2;
    Line() {}
    Line(Point p1, Point p2) : p1(p1), p2(p2) {}
};

double dis(Point A, Point B) {
    return sqrt((A.x - B.x) * (A.x - B.x) + (A.y - B.y) * (A.y - B.y));
}

int dcmp(double x) {
    if (fabs(x) < eps) return 0;
    return x < 0 ? -1 : 1;
}

Point getIntersection(Line a, Line b) {
    double x1 = a.p1.x, y1 = a.p1.y;
    double x2 = a.p2.x, y2 = a.p2.y;
    double x3 = b.p1.x, y3 = b.p1.y;
    double x4 = b.p2.x, y4 = b.p2.y;
    double k1 = (y2 - y1) / (x2 - x1);
    double k2 = (y4 - y3) / (x4 - x3);
    double b1 = y1 - k1 * x1;
    double b2 = y3 - k2 * x3;
    double x = (b2 - b1) / (k1 - k2);
    double y = k1 * x + b1;
    return Point(x, y);
}

void getOuterTangents(Circle A, Circle B, Line &l1, Line &l2) {
    double d = dis(A.c, B.c);
    double cosA = (A.r - B.r) / d;
    double sinA = sqrt(1 - cosA * cosA);
    Point v = (B.c - A.c) / d;
    Point p1 = A.c + v * (A.r / cosA);
    Point p2 = B.c - v * (B.r / cosA);
    Point p3 = A.c + v * (A.r / cosA * cosA * sinA);
    Point p4 = B.c - v * (B.r / cosA * cosA * sinA);
    l1 = Line(p1, p2);
    l2 = Line(p3, p4);
}

int main() {
    Circle A(Point(0, 0), 2);
    Circle B(Point(3, 0), 1);
    Line l1, l2;
    getOuterTangents(A, B, l1, l2);
    cout << "Outer tangents: " << endl;
    cout << "(" << l1.p1.x << ", " << l1.p1.y << ") -> (" << l1.p2.x << ", " << l1.p2.y << ")" << endl;
    cout << "(" << l2.p1.x << ", " << l2.p1.y << ") -> (" << l2.p2.x << ", " << l2.p2.y << ")" << endl;
    return 0;
}

这个例子中，我们假设两个圆的圆心分别为 (0, 0) 和 (3, 0)，半径分别为 2 和 1。程序输出了两个圆的外公切线的坐标。在实际使用中，可以根据需要进行修改。