## 1) Abridged problem statement

For each test case, you are given a circle with center \((X_0,Y_0)\) and radius \(R\), and two points \(A\) and \(B\) strictly inside the circle. Find any point \(C\) on the circle such that the angle \(\angle ACB\) is as large as possible. Output \(C\) with 6 digits after the decimal point.

---

## 2) Detailed editorial (geometry + derivation)

### Key observation: maximizing \(\angle ACB\) ↔ minimizing circumradius of \(\triangle ABC\)
For any non-degenerate triangle \(ABC\), the circumradius \(\rho\) satisfies the inscribed-angle / chord relation:
\[
\sin(\angle ACB) \;=\; \frac{|AB|}{2\rho}.
\]
Here \(|AB|\) is fixed, so maximizing \(\angle ACB\) (equivalently maximizing \(\sin(\angle ACB)\) for angles in \((0,\pi)\)) is equivalent to **minimizing the circumradius \(\rho\)** of the circle passing through \(A,B,C\).

So our problem becomes:

> Choose \(C\) on the given circle such that the circle through \(A,B,C\) has minimum possible radius.

### Locus of circumcenters of circles through \(A\) and \(B\)
Consider all circles passing through fixed points \(A\) and \(B\). Their centers lie on the **perpendicular bisector** of segment \(AB\).

Work in coordinates translated so the given circle’s center is at the origin:
- Let \(O=(0,0)\) after translation.
- Replace \(A,B\) by \(A',B'\) where \(A'=A-O\), \(B'=B-O\). (The code calls them `Al, Bl`.)

Define:
- Midpoint \(M = \frac{A'+B'}{2}\).
- A vector perpendicular to \(AB\): \(V = (B'-A')^\perp\). Note \(|V| = |AB| = D\).

Every center \(P\) of a circle through \(A'\) and \(B'\) can be written as:
\[
P(t) = M + tV, \quad t \in \mathbb{R}.
\]

Its radius \(\rho(t)\) is distance from \(P(t)\) to \(A'\):
\[
\rho(t)^2 = |P(t)-A'|^2.
\]
Since \(M-A'\) is along \(AB\) and \(V\) is perpendicular to \(AB\), they are orthogonal, so:
- \(|M-A'| = D/2\)
- \(|tV| = |t|D\)

Thus:
\[
\rho(t)^2 = (D/2)^2 + (tD)^2 = D^2\left(t^2+\frac14\right)
\]
\[
\Rightarrow \rho(t) = D\sqrt{t^2+\frac14}.
\]

### Why the optimum happens at tangency
We are allowed only those circles (center \(P(t)\), radius \(\rho(t)\)) that intersect the given circle \(|X|=R\) (because \(C\) must lie on that given circle and also on the circle through \(A,B\)).

As we slide \(t\), the circle through \(A,B\) changes continuously, hence \(\rho(t)\) changes continuously. Among all feasible \(t\) values where intersection exists, the minimum radius occurs at the **boundary** of feasibility, i.e. when the two circles are **tangent** (intersection degenerates to one point). This is a standard “minimum under intersection constraint” argument: if the circles intersect in two points, you can slightly move toward reducing \(\rho\) until they become tangent.

So we only need to consider circles through \(A,B\) that are tangent to the given circle.

### Tangency condition → quadratic in \(t\)
Tangency between:
- given circle centered at \(O\) radius \(R\)
- circle centered at \(P(t)\) radius \(\rho(t)\)

means the distance between centers equals \(R \pm \rho(t)\):
\[
|P(t)| = R + \rho(t) \quad \text{(external tangency)}
\]
or
\[
|P(t)| = |R - \rho(t)| \quad \text{(internal tangency)}.
\]

Square and eliminate the square root carefully; the provided code uses a derived quadratic of the form:
\[
(\,b^2 - 4R^2D^2\,)t^2 + 2bqt + (q^2 - R^2D^2) = 0
\]
where
- \(b = 2(M\cdot V)\)
- \(q = |M|^2 - R^2 - \frac{D^2}{4}\)

Solving this quadratic yields up to two real candidates \(t\) (some cases yield one due to degeneracy).

### Recover candidate \(C\) from \(t\)
For each root \(t\):
1. Compute center of the “\(A,B\)-circle”: \(P = M + tV\).
2. Compute its radius \(\rho = D\sqrt{t^2+\frac14}\).
3. The tangency point \(C\) lies on the ray from \(O\) to \(P\), at distance \(R\) from \(O\):
   \[
   C = \pm R \frac{P}{|P|}.
   \]
   There are two antipodal points on the given circle; only one will satisfy that \(C\) is actually at distance \(\rho\) from \(P\) (within numeric tolerance):
   \[
   |C-P| \approx \rho.
   \]
4. For valid candidates, compute \(\cos(\angle ACB)\). Since \(\angle\) is maximized when \(\cos\) is minimized (angle in \([0,\pi]\)):
   \[
   \cos(\angle ACB) = \frac{(A'-C)\cdot(B'-C)}{|A'-C||B'-C|}.
   \]
   Choose the candidate with smallest cosine.

Finally, translate back by adding the original circle center \(O\).

### Complexity
Each test case solves a quadratic and checks a constant number of candidates: **\(O(1)\)** time. This is crucial for up to 10,000 tests.

---

## 3) Provided C++ solution with detailed line-by-line comments

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

// Print a pair as "first second"
template<typename T1, typename T2>
ostream& operator<<(ostream& out, const pair<T1, T2>& x) {
    return out << x.first << ' ' << x.second;
}

// Read a pair as "first second"
template<typename T1, typename T2>
istream& operator>>(istream& in, pair<T1, T2>& x) {
    return in >> x.first >> x.second;
}

// Read a vector of values
template<typename T>
istream& operator>>(istream& in, vector<T>& a) {
    for (auto& x : a) in >> x;
    return in;
}

// Print a vector (space-separated)
template<typename T>
ostream& operator<<(ostream& out, const vector<T>& a) {
    for (auto x : a) out << x << ' ';
    return out;
}

// Use double for coordinates
using coord_t = double;

struct Point {
    // Epsilon for floating comparisons
    static constexpr coord_t eps = 1e-9;
    // PI constant
    static inline const coord_t PI = acos((coord_t)-1.0);

    coord_t x, y;
    Point(coord_t x = 0, coord_t y = 0) : x(x), y(y) {}

    // Vector addition/subtraction/scalar ops
    Point operator+(const Point& p) const { return Point(x + p.x, y + p.y); }
    Point operator-(const Point& p) const { return Point(x - p.x, y - p.y); }
    Point operator*(coord_t c) const { return Point(x * c, y * c); }
    Point operator/(coord_t c) const { return Point(x / c, y / c); }

    // Dot product
    coord_t operator*(const Point& p) const { return x * p.x + y * p.y; }
    // Cross product (2D scalar cross)
    coord_t operator^(const Point& p) const { return x * p.y - y * p.x; }

    // Comparisons (exact; mainly unused for geometry decisions)
    bool operator==(const Point& p) const { return x == p.x && y == p.y; }
    bool operator!=(const Point& p) const { return x != p.x || y != p.y; }
    bool operator<(const Point& p) const {
        return x != p.x ? x < p.x : y < p.y;
    }
    bool operator>(const Point& p) const {
        return x != p.x ? x > p.x : y > p.y;
    }
    bool operator<=(const Point& p) const {
        return x != p.x ? x < p.x : y <= p.y;
    }
    bool operator>=(const Point& p) const {
        return x != p.x ? x > p.x : y >= p.y;
    }

    // Squared length and length
    coord_t norm2() const { return x * x + y * y; }
    coord_t norm() const { return sqrt(norm2()); }

    // Polar angle
    coord_t angle() const { return atan2(y, x); }

    // Rotate by angle a (radians) around origin
    Point rotate(coord_t a) const {
        return Point(x * cos(a) - y * sin(a), x * sin(a) + y * cos(a));
    }

    // Perpendicular vector (rotated +90 degrees)
    Point perp() const { return Point(-y, x); }

    // Unit vector (normalized)
    Point unit() const { return *this / norm(); }

    // Unit normal (perpendicular then normalized)
    Point normal() const { return perp().unit(); }

    // Project point p onto this vector (treated as direction from origin)
    Point project(const Point& p) const {
        return *this * (*this * p) / norm2();
    }

    // Reflect point p across the line through origin in direction of this vector
    Point reflect(const Point& p) const {
        return *this * 2 * (*this * p) / norm2() - p;
    }

    // Stream operators for Point
    friend ostream& operator<<(ostream& os, const Point& p) {
        return os << p.x << ' ' << p.y;
    }
    friend istream& operator>>(istream& is, Point& p) {
        return is >> p.x >> p.y;
    }

    // ccw test: returns +1,0,-1 depending on orientation of triangle (a,b,c)
    friend int ccw(const Point& a, const Point& b, const Point& c) {
        coord_t v = (b - a) ^ (c - a);
        if (-eps <= v && v <= eps) return 0;
        else if (v > 0) return 1;
        else return -1;
    }

    // Check if p lies on segment ab (with epsilon)
    friend bool point_on_segment(const Point& a, const Point& b, const Point& p) {
        return ccw(a, b, p) == 0 &&
               p.x >= min(a.x, b.x) - eps && p.x <= max(a.x, b.x) + eps &&
               p.y >= min(a.y, b.y) - eps && p.y <= max(a.y, b.y) + eps;
    }

    // Check if p is inside or on boundary of triangle abc
    friend bool point_in_triangle(
        const Point& a, const Point& b, const Point& c, const Point& p
    ) {
        int d1 = ccw(a, b, p);
        int d2 = ccw(b, c, p);
        int d3 = ccw(c, a, p);
        return (d1 >= 0 && d2 >= 0 && d3 >= 0) ||
               (d1 <= 0 && d2 <= 0 && d3 <= 0);
    }

    // Intersection point of infinite lines a1-b1 and a2-b2 (assumes not parallel)
    friend Point line_line_intersection(
        const Point& a1, const Point& b1, const Point& a2, const Point& b2
    ) {
        return a1 +
               (b1 - a1) * ((a2 - a1) ^ (b2 - a2)) / ((b1 - a1) ^ (b2 - a2));
    }

    // Are vectors a and b collinear (cross close to 0)?
    friend bool collinear(const Point& a, const Point& b) {
        return abs(a ^ b) < eps;
    }

    // Circumcenter of triangle abc (intersection of perpendicular bisectors)
    friend Point circumcenter(const Point& a, const Point& b, const Point& c) {
        Point mid_ab = (a + b) / 2.0;
        Point mid_ac = (a + c) / 2.0;
        Point perp_ab = (b - a).perp();
        Point perp_ac = (c - a).perp();
        return line_line_intersection(
            mid_ab, mid_ab + perp_ab, mid_ac, mid_ac + perp_ac
        );
    }

    // Signed area contribution of circular arc sector (unused in this solution)
    friend coord_t arc_area(
        const Point& center, coord_t r, const Point& p1, const Point& p2
    ) {
        coord_t theta1 = (p1 - center).angle();
        coord_t theta2 = (p2 - center).angle();
        if (theta2 < theta1 - eps) theta2 += 2 * PI;

        coord_t d_theta = theta2 - theta1;
        coord_t cx = center.x, cy = center.y;
        coord_t area = r * cx * (sin(theta2) - sin(theta1)) -
                       r * cy * (cos(theta2) - cos(theta1)) + r * r * d_theta;
        return area / 2.0;
    }

    // Intersection points of circles (unused in this solution)
    friend vector<Point> intersect_circles(
        const Point& c1, coord_t r1, const Point& c2, coord_t r2
    ) {
        Point d = c2 - c1;
        coord_t dist = d.norm();

        // No intersection if too far, one inside another, or same center
        if (dist > r1 + r2 + eps || dist < abs(r1 - r2) - eps || dist < eps) {
            return {};
        }

        // Standard circle-circle intersection formula
        coord_t a = (r1 * r1 - r2 * r2 + dist * dist) / (2 * dist);
        coord_t h_sq = r1 * r1 - a * a;
        if (h_sq < -eps) return {};
        if (h_sq < 0) h_sq = 0;
        coord_t h = sqrt(h_sq);

        Point mid = c1 + d.unit() * a;
        Point perp_dir = d.perp().unit();

        // Tangent => one intersection
        if (h < eps) return {mid};

        // Two intersections
        return {mid + perp_dir * h, mid - perp_dir * h};
    }

    // Intersection of a ray and a segment (unused in this solution)
    friend optional<Point> intersect_ray_segment(
        const Point& ray_start, const Point& ray_through,
        const Point& seg_a, const Point& seg_b
    ) {
        Point ray_dir = ray_through - ray_start;
        if (ray_dir.norm2() < Point::eps) return {}; // degenerate ray

        Point seg_dir = seg_b - seg_a;
        coord_t denom = ray_dir ^ seg_dir;
        if (fabs(denom) < eps) return {}; // parallel

        // Solve ray_start + t*ray_dir = seg_a + s*seg_dir
        coord_t t = ((seg_a - ray_start) ^ seg_dir) / denom;
        if (t < eps) return {}; // intersection behind ray start

        coord_t s = ((seg_a - ray_start) ^ ray_dir) / denom;
        if (s < eps || s > 1 - eps) return {}; // outside segment interior (epsilon)
        return ray_start + ray_dir * t;
    }
};

Point cen, A, B; // circle center, points A and B (global per test)
coord_t R;       // radius

// Read one test
void read() { cin >> cen >> R >> A >> B; }

void solve() {
    // Translate so circle center is at origin
    Point Oshift = cen;
    Point Al = A - Oshift, Bl = B - Oshift;

    // M = midpoint of A and B in shifted coordinates
    Point M = (Al + Bl) / 2.0;

    // V = perpendicular to (B-A). This spans the perpendicular bisector direction.
    Point V = (Bl - Al).perp();

    // D = |AB|
    coord_t D = (Bl - Al).norm();
    coord_t Dsq = D * D;

    // Derived constants for the quadratic in t
    coord_t b = 2.0 * (M * V);               // 2 * (M dot V)
    coord_t q = M.norm2() - R * R - Dsq / 4.0;

    // Quadratic coefficients: qa*t^2 + qb*t + qc = 0
    coord_t qa = b * b - 4.0 * R * R * Dsq;
    coord_t qb = 2.0 * b * q;
    coord_t qc = q * q - R * R * Dsq;

    // Solve for candidate t values (tangency positions)
    vector<coord_t> ts;
    if (fabs(qa) < 1e-18) {
        // Degenerates to linear equation qb*t + qc = 0
        if (fabs(qb) > 1e-18) {
            ts.push_back(-qc / qb);
        }
    } else {
        // Quadratic formula
        coord_t disc = qb * qb - 4.0 * qa * qc;
        if (disc < 0) disc = 0;          // clamp due to floating errors
        coord_t sq = sqrt(disc);
        ts.push_back((-qb + sq) / (2.0 * qa));
        ts.push_back((-qb - sq) / (2.0 * qa));
    }

    // Among all valid tangency candidates, pick the one maximizing angle ACB
    // by minimizing cos(angle).
    Point best_C;
    coord_t best_cos = 2.0; // cos ranges [-1,1], so 2 is "worst" initial

    for (coord_t t : ts) {
        // Center of circle through A and B at parameter t
        Point P = M + V * t;

        // Radius of that circle: r = D * sqrt(t^2 + 1/4)
        coord_t r = D * sqrt(t * t + 0.25);

        // If P is at origin, direction to tangency undefined
        coord_t pnorm = P.norm();
        if (pnorm < Point::eps) continue;

        // Tangency point on given circle lies on line OP; try both directions
        for (int sign : {1, -1}) {
            // Candidate point C on given circle (radius R)
            Point C = P * (sign * R / pnorm);

            // Verify that C is also on the circle centered at P with radius r
            if (fabs((C - P).norm() - r) > 1e-6) continue;

            // Compute cos(angle ACB) using vectors from C to A and B
            Point ua = Al - C, ub = Bl - C;
            coord_t na = ua.norm(), nb = ub.norm();
            if (na < Point::eps || nb < Point::eps) continue; // avoid degenerate

            coord_t cosv = (ua * ub) / (na * nb);

            // Larger angle => smaller cosine
            if (cosv < best_cos) {
                best_cos = cosv;
                best_C = C;
            }
        }
    }

    // Translate answer back to original coordinates
    Point ans = best_C + Oshift;

    // Output with required precision
    cout << fixed << setprecision(6) << ans.x << ' ' << ans.y << '\n';
}

int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);

    int T = 1;
    cin >> T;
    for (int test = 1; test <= T; test++) {
        read();
        solve();
    }
    return 0;
}
```

---

## 4) Python solution (same approach) with detailed comments

```python
import sys
import math

EPS = 1e-9

class Point:
    __slots__ = ("x", "y")
    def __init__(self, x=0.0, y=0.0):
        self.x = float(x)
        self.y = float(y)

    # vector ops
    def __add__(self, other): return Point(self.x + other.x, self.y + other.y)
    def __sub__(self, other): return Point(self.x - other.x, self.y - other.y)
    def __mul__(self, k):     return Point(self.x * k, self.y * k)  # scalar
    def __truediv__(self, k): return Point(self.x / k, self.y / k)

    # dot and cross
    def dot(self, other):   return self.x * other.x + self.y * other.y
    def cross(self, other): return self.x * other.y - self.y * other.x

    def norm2(self): return self.x * self.x + self.y * self.y
    def norm(self):  return math.sqrt(self.norm2())

    def perp(self):
        # rotate by +90 degrees
        return Point(-self.y, self.x)


def solve_case(x0, y0, R, xa, ya, xb, yb):
    # Shift coordinates so circle center becomes origin
    O = Point(x0, y0)
    A = Point(xa, ya) - O
    B = Point(xb, yb) - O
    R = float(R)

    # Midpoint of AB
    M = (A + B) / 2.0

    # Perpendicular direction to AB (not normalized)
    V = (B - A).perp()

    # D = |AB|
    D = (B - A).norm()
    Dsq = D * D

    # Quadratic coefficients as in the C++ solution
    b = 2.0 * M.dot(V)
    q = M.norm2() - R * R - Dsq / 4.0

    qa = b * b - 4.0 * R * R * Dsq
    qb = 2.0 * b * q
    qc = q * q - R * R * Dsq

    ts = []
    if abs(qa) < 1e-18:
        # linear qb*t + qc = 0
        if abs(qb) > 1e-18:
            ts.append(-qc / qb)
    else:
        disc = qb * qb - 4.0 * qa * qc
        if disc < 0.0:
            disc = 0.0
        sq = math.sqrt(disc)
        ts.append((-qb + sq) / (2.0 * qa))
        ts.append((-qb - sq) / (2.0 * qa))

    best_cos = 2.0
    best_C = None

    for t in ts:
        # Center P of the circle through A and B
        P = M + (V * t)

        # Radius of that circle
        r = D * math.sqrt(t * t + 0.25)

        pnorm = P.norm()
        if pnorm < EPS:
            continue

        # Try both points on the given circle along the line OP
        for sign in (1.0, -1.0):
            C = P * (sign * R / pnorm)

            # Must also lie on circle centered at P with radius r (tangency)
            if abs((C - P).norm() - r) > 1e-6:
                continue

            ua = A - C
            ub = B - C
            na = ua.norm()
            nb = ub.norm()
            if na < EPS or nb < EPS:
                continue

            cosv = ua.dot(ub) / (na * nb)
            if cosv < best_cos:
                best_cos = cosv
                best_C = C

    # Translate back to original coordinates
    ans = best_C + O
    return ans.x, ans.y


def main():
    data = sys.stdin.buffer.read().split()
    t = int(data[0])
    idx = 1
    out_lines = []
    for _ in range(t):
        x0 = int(data[idx]); y0 = int(data[idx+1]); R = int(data[idx+2])
        xa = int(data[idx+3]); ya = int(data[idx+4])
        xb = int(data[idx+5]); yb = int(data[idx+6])
        idx += 7

        x, y = solve_case(x0, y0, R, xa, ya, xb, yb)
        out_lines.append(f"{x:.6f} {y:.6f}")

    sys.stdout.write("\n".join(out_lines))

if __name__ == "__main__":
    main()
```

---

## 5) Compressed editorial

Translate so the circle center is at the origin. For any point \(C\) on the circle, \(\angle ACB\) is maximized iff the circumradius \(\rho\) of \(\triangle ABC\) is minimized because \(\sin(\angle ACB)=|AB|/(2\rho)\). Centers of circles through \(A,B\) lie on the perpendicular bisector: \(P(t)=M+tV\), where \(M=(A+B)/2\), \(V=(B-A)^\perp\), \(D=|AB|\), and \(\rho(t)=D\sqrt{t^2+1/4}\). Feasible \(t\) require intersection with the given circle \(|X|=R\); the minimum \(\rho\) over feasible \(t\) occurs on the boundary, i.e. when the two circles are tangent. Tangency yields a quadratic in \(t\):
\((b^2-4R^2D^2)t^2+2bqt+(q^2-R^2D^2)=0\),
with \(b=2(M\cdot V)\), \(q=|M|^2-R^2-D^2/4\). For each real root, compute \(P,\rho\), then tangency point \(C=\pm R\,P/|P|\) and keep the one satisfying \(|C-P|\approx \rho\). Among candidates choose the one with minimal \(\cos(\angle ACB)\). Output \(C\) (shifted back).