## 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) C++ Solution

```cpp
#include <bits/stdc++.h>
// #include <coding_library/geometry/point.hpp>

using namespace std;

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

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

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

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

using coord_t = double;

struct Point {
    static constexpr coord_t eps = 1e-9;
    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) {}

    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); }

    coord_t operator*(const Point& p) const { return x * p.x + y * p.y; }
    coord_t operator^(const Point& p) const { return x * p.y - y * p.x; }

    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;
    }

    coord_t norm2() const { return x * x + y * y; }
    coord_t norm() const { return sqrt(norm2()); }
    coord_t angle() const { return atan2(y, x); }

    Point rotate(coord_t a) const {
        return Point(x * cos(a) - y * sin(a), x * sin(a) + y * cos(a));
    }

    Point perp() const { return Point(-y, x); }
    Point unit() const { return *this / norm(); }
    Point normal() const { return perp().unit(); }
    Point project(const Point& p) const {
        return *this * (*this * p) / norm2();
    }
    Point reflect(const Point& p) const {
        return *this * 2 * (*this * p) / norm2() - p;
    }

    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;
    }

    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;
        }
    }

    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;
    }

    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);
    }

    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));
    }

    friend bool collinear(const Point& a, const Point& b) {
        return abs(a ^ b) < eps;
    }

    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
        );
    }

    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;
    }

    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();

        if(dist > r1 + r2 + eps || dist < abs(r1 - r2) - eps || dist < eps) {
            return {};
        }

        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();

        if(h < eps) {
            return {mid};
        }
        return {mid + perp_dir * h, mid - perp_dir * h};
    }

    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 {};
        }
        Point seg_dir = seg_b - seg_a;
        coord_t denom = ray_dir ^ seg_dir;
        if(fabs(denom) < eps) {
            return {};
        }
        coord_t t = ((seg_a - ray_start) ^ seg_dir) / denom;
        if(t < eps) {
            return {};
        }
        coord_t s = ((seg_a - ray_start) ^ ray_dir) / denom;
        if(s < eps || s > 1 - eps) {
            return {};
        }
        return ray_start + ray_dir * t;
    }
};

Point cen, A, B;
coord_t R;

void read() { cin >> cen >> R >> A >> B; }

void solve() {
    // One way to solve this is to split it into two cases - point C is "below"
    // the line AB, and another one for above. Then do a ternary search. There
    // is also a closed form solution which we describe here. By the inscribed
    // angle theorem, if C lies on a circle through A and B of radius r, then
    // sin(ACB) = |AB| / (2r). So for a fixed chord AB, the angle ACB depends
    // only on the radius of the circle through A, B, C, and on which side of AB
    // the point C lies. Because we have sin(.), we want to maximize it, or we
    // want to minimize the radius of the ACB circle. Among circles through A
    // and B that intersect the given circle, sliding the center along the
    // perpendicular bisector of AB changes the radius continuously; the
    // extremal values of the inscribed angle at the intersection point occur
    // exactly when the circle through A, B, C is tangent to the given circle.
    // So the candidates for C are the (at most two) tangency points of circles
    // through A, B with the given circle.
    //
    // Translate so that the center of the given circle is at the origin. Let
    // M = (A + B) / 2, V = (B - A).perp() so that |V| = |AB| = D. Centers of
    // circles through A and B are parameterized by P(t) = M + t V, with
    // radius:
    //
    //     r(t) = D * sqrt(t^2 + 1/4)
    //
    // (since |P(t) - A|^2 = D^2/4 + t^2 D^2 as M - A is perpendicular to V).
    // Tangency with the given circle is equivalent to |OP(t)| = |R +/- r(t)|.
    // Writing |OP(t)|^2 = |M|^2 + 2 t (M . V) + t^2 D^2 and squaring after
    // isolating the radical gives the quadratic
    //
    //     (b^2 - 4 R^2 D^2) t^2 + 2 b q t + (q^2 - R^2 D^2) = 0,
    //
    // where b = 2 (M . V) and q = |M|^2 - R^2 - D^2/4. For each real root t
    // compute P = M + t V and r = D sqrt(t^2 + 1/4); the tangency point lies
    // on line OP at distance R from O, so it is one of +/- R * P/|P|. Keep
    // the one that satisfies |C - P| ~= r, and among the valid candidates
    // pick the one with the smallest cos(ACB).

    Point Oshift = cen;
    Point Al = A - Oshift, Bl = B - Oshift;

    Point M = (Al + Bl) / 2.0;
    Point V = (Bl - Al).perp();
    coord_t D = (Bl - Al).norm();
    coord_t Dsq = D * D;

    coord_t b = 2.0 * (M * V);
    coord_t q = M.norm2() - R * R - Dsq / 4.0;

    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;

    vector<coord_t> ts;
    if(fabs(qa) < 1e-18) {
        if(fabs(qb) > 1e-18) {
            ts.push_back(-qc / qb);
        }
    } else {
        coord_t disc = qb * qb - 4.0 * qa * qc;
        if(disc < 0) {
            disc = 0;
        }
        coord_t sq = sqrt(disc);
        ts.push_back((-qb + sq) / (2.0 * qa));
        ts.push_back((-qb - sq) / (2.0 * qa));
    }

    Point best_C;
    coord_t best_cos = 2.0;
    for(coord_t t: ts) {
        Point P = M + V * t;
        coord_t r = D * sqrt(t * t + 0.25);
        coord_t pnorm = P.norm();
        if(pnorm < Point::eps) {
            continue;
        }
        for(int sign: {1, -1}) {
            Point C = P * (sign * R / pnorm);
            if(fabs((C - P).norm() - r) > 1e-6) {
                continue;
            }
            Point ua = Al - C, ub = Bl - C;
            coord_t na = ua.norm(), nb = ub.norm();
            if(na < Point::eps || nb < Point::eps) {
                continue;
            }
            coord_t cosv = (ua * ub) / (na * nb);
            if(cosv < best_cos) {
                best_cos = cosv;
                best_C = C;
            }
        }
    }

    Point ans = best_C + Oshift;
    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).