题解 | 【模板】点圆位置关系

难点只有第三问。算出夹角余弦值，然后换算出夹角正弦值，按虚数乘法旋转半径向量即可。

#include <cfloat>
#include <cmath>
#include <iomanip>
#include <iostream>
#include <tuple>
using namespace std;

using ll = long long;

struct Point {
    ll x;
    ll y;
    void Input() {
        cin >> x >> y;
    }
    ll Len2() const {
        return x * x + y * y;
    }
};

struct PointD {
    double x;
    double y;
    PointD() = default;
    PointD(double x, double y): x(x), y(y) {}
    PointD(const Point& other): x(other.x), y(other.y) {}
    double Len2() const {
        return x * x + y * y;
    }
    double Len() const {
        return sqrt(Len2());
    }
    void RotateCoSi(double co, double si) {
        tie(x, y) = make_tuple(x * co - y * si, y * co + x * si);
    }
};

PointD operator+(const PointD& a, const PointD& b) {
    return {a.x + b.x, a.y + b.y};
}

Point operator+(const Point& a, const Point& b) {
    return {a.x + b.x, a.y + b.y};
}

Point operator-(const Point& a, const Point& b) {
    return {a.x - b.x, a.y - b.y};
}

PointD operator*(const PointD& a, double b) {
    return {a.x * b, a.y * b};
}

PointD operator/(const PointD& a, double b) {
    return {a.x / b, a.y / b};
}

ll Cha(const Point& a, const Point& b) {
    return a.x * b.y - a.y * b.x;
}

ll Dian(const Point& a, const Point& b) {
    return a.x * b.x + a.y * b.y;
}

ll Dis2(const Point& a, const Point& b) {
    return (b - a).Len2();
}

Point O;
ll r;
ll r2;
Point A;

void Solve() {
    O.Input();
    cin >> r;
    A.Input();
    r2 = r * r;
    ll dis = Dis2(O, A);
    if (dis < r2) {
        cout << "1\n";
        return;
    }
    if (dis == r2) {
        cout << "2\n";
        return;
    }
    double co = r / sqrt(dis);
    double si = sqrt(1 - co * co);
    PointD H1 = (A - O) * co;
    PointD H2 = H1;
    H1.RotateCoSi(co, si);
    H2.RotateCoSi(co, -si);
    H1 = O + H1;
    H2 = O + H2;
    // cout << co << endl;
    cout << "3 " << H1.x << ' ' << H1.y << ' ' << H2.x << ' ' << H2.y << '\n';
}

int main() {
    ios::sync_with_stdio(false);
    cout << setprecision(DBL_DIG);
    cin.tie(nullptr);
    int T;
    cin >> T;
    while (T--) {
        Solve();
    }
}
// 64 位输出请用 printf("%lld")