#define _CRT_SEstartE_NO_DEPRECATE
#pragma warning(disable : 4996)
#include <map>
#include <unordered_map>
#include <set>
#include <fstream>
#include <queue>
#include <deque>
#include <stack>
#include <vector>
#include <string>
#include <iostream>
#include <algorithm>
#include <iterator>
#include <bitset>
#include <cstring>
#include <cstdio>
#include <cmath>
#include <cstdio>
#include <bits/stdc++.h>
#define ACcode ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0);
typedef long long ll;
typedef unsigned long long ull;
const ll maxn = 1e6 + 7;
const ll maxm = 7e6 + 7;
constexpr ll mod = 998244353;
const ll inf = 0x3f3f3f3f;
const ll INF = 0x3f3f3f3f3f3f3f3f;
const int Prime = 100007;
const double eps = 1e-10;
const double pi = acos(-1.0);
using namespace std;

using point_t = long long;
template <typename T>
struct point
{
    T x, y;
    point() {}
    point(T _x, T _y) { x = _x, y = _y; }
    point operator+(const point &b) const { return {x + b.x, y + b.y}; }
    point operator-(const point &b) const { return {x - b.x, y - b.y}; }
    point operator/(const T &k) const { return {x / k, y / k}; }
    point operator*(const T &k) const { return {x * k, y * k}; }
    T operator*(const point &b) const { return (x * b.x + y * b.y); }
    T operator^(const point &b) const { return x * b.y - y * b.x; }
    // 浮点数运算
    T len2() const { return (*this) * (*this); }
    T dis2(const point &a) const { return (a - (*this)).len2(); }
    long double len() const { return sqrtl(len2()); }
    long double dis(const point &a) const { return sqrtl(dis2(a)); }
};
using Point = point<point_t>;
template <typename T>
struct line
{
    point<T> p, v; // 直线上一点和方向向量
    line() {}
    line(T k, T b) { p = {0, b}, v = {1, k}; }
    line(point<T> _p, point<T> _v) { p = _p, v = _v; }
    point<T> proj(const point<T> &a) const // 点在直线上的投影
    {
        return p + v * ((v * (a - p)) / (v * v));
    }
    long double dis(const point<T> &a) const // 点到直线距离
    {
        return abs(v ^ (a - p)) / v.len();
    }
};
using Line = line<point_t>;
struct Circle
{
    Point c;
    long double r;
    // 点与圆的关系
    // -1 圆上 | 0 圆外 | 1 圆内
    int is_in(const Point &p) const
    {
        const long double d = p.dis(c);
        return abs(d - r) <= eps ? -1 : d < r - eps;
    }
    // 直线与圆关系
    // 0 相离 | 1 相切 | 2 相交
    int relation(const Line &l) const
    {
        const long double d = l.dis(c);
        if (d > r + eps)
            return 0;
        if (abs(d - r) <= eps)
            return 1;
        return 2;
    }
    // 直线与圆的交点
    vector<Point> inter(const Line &l) const
    {
        const long double d = l.dis(c);
        const Point p = l.proj(c);
        const int t = relation(l);
        if (t == 0)
            return vector<Point>();
        if (t == 1)
            return vector<Point>{p};
        const long double k = sqrt(r * r - d * d);
        return vector<Point>{p - (l.v / l.v.len()) * k, p + (l.v / l.v.len()) * k};
    }
};

void solve()
{
    cout << fixed << setprecision(0);
    Point pc;
    Circle cir;
    cin >> cir.c.x >> cir.c.y >> cir.r >> pc.x >> pc.y;
    if (cir.is_in(pc) != 0)
    {
        cout << 0 << '\n';
        cout << pc.x << " " << pc.y << '\n';
        return;
    }
    Line l1(pc, {pc.x - cir.c.x, pc.y - cir.c.y});
    vector<Point> temp = cir.inter(l1);
    Point cro;
    if (pc.dis2(temp[0]) < pc.dis2(temp[1]))
        cro = temp[0];
    else
        cro = temp[1];
    Point ans;
    ll ansdis = 4e18;
    for (ll x = cro.x - 1200; x <= cro.x + 1200; x++)
    {
        for (ll y = cro.y - 1200; y <= cro.y + 1200; y++)
        {
            Point now(x, y);
            if (cir.is_in(now) != 0)
            {
                ll nowdis = pc.dis2(now);
                if (nowdis < ansdis)
                {
                    ans = now;
                    ansdis = nowdis;
                }
            }
        }
    }
    cout << 1 << '\n';
    cout << pc.x << " " << pc.y << " " << ans.x << " " << ans.y << '\n';
}

signed main()
{
    ACcode;
    // freopen("house.in", "r", stdin);
    // freopen("house.out", "w", stdout);
    int t = 1;
    cin >> t;
    while (t--)
    {
        solve();
    }
    return 0;
}