Let one of the vertices of the square circumseribing the circle `x^(2) + y^(2) - 6x -4y + 11 = 0` be `(4, 2 + sqrt(3))` The other vertices of the square may be

Options

A. `(3 -sqrt(3), 3)`
B. `(2, 2 -sqrt(3))`
C. `(3+ sqrt(3 ). 1)`
D. `(0, 0)`

Answer

To find the other vertices of the square circumscribing the given circle, we will follow a systematic approach. Let's break it down step by step.

### Step 1: Find the center and radius of the circle
The equation of the circle is given as:
\[ x^2 + y^2 - 6x - 4y + 11 = 0 \]

We can rewrite this in standard form by completing the square.

1. Rearranging the equation:
\[ (x^2 - 6x) + (y^2 - 4y) + 11 = 0 \]

2. Completing the square for \(x\) and \(y\):
- For \(x^2 - 6x\), add and subtract \(9\) (which is \((\frac{6}{2})^2\)):
\[ (x - 3)^2 - 9 \]
- For \(y^2 - 4y\), add and subtract \(4\) (which is \((\frac{4}{2})^2\)):
\[ (y - 2)^2 - 4 \]

3. Substitute back:
\[ (x - 3)^2 - 9 + (y - 2)^2 - 4 + 11 = 0 \]
\[ (x - 3)^2 + (y - 2)^2 - 2 = 0 \]
\[ (x - 3)^2 + (y - 2)^2 = 2 \]

From this, we can see that the center of the circle \(C\) is \((3, 2)\) and the radius \(r\) is \(\sqrt{2}\).

### Step 2: Find the coordinates of the given vertex
The given vertex of the square is:
\[ P(4, 2 + \sqrt{3}) \]

### Step 3: Calculate the distance from the center to the vertex
Using the distance formula, we find the distance \(OP\):
\[ OP = \sqrt{(4 - 3)^2 + ((2 + \sqrt{3}) - 2)^2} = \sqrt{1^2 + (\sqrt{3})^2} = \sqrt{1 + 3} = \sqrt{4} = 2 \]

### Step 4: Determine the angle of the line from center to vertex
To find the slope of line \(OP\):
\[ \text{slope} = \frac{(2 + \sqrt{3}) - 2}{4 - 3} = \sqrt{3} \]
This corresponds to an angle \(\theta\) where:
\[ \tan(\theta) = \sqrt{3} \implies \theta = 60^\circ \]

### Step 5: Calculate the coordinates of the other vertices
Since the square's diagonals bisect each other at right angles, we can find the other vertices by rotating the point \(P\) around the center \(C\) by \(90^\circ\) and \(270^\circ\).

1. **Vertex Q (90° rotation)**:
- Using rotation formulas:
\[ Q_x = C_x + (P_x - C_x) \cdot \cos(90^\circ) - (P_y - C_y) \cdot \sin(90^\circ) \]
\[ Q_y = C_y + (P_x - C_x) \cdot \sin(90^\circ) + (P_y - C_y) \cdot \cos(90^\circ) \]
Substituting values:
\[ Q_x = 3 + (4 - 3) \cdot 0 - ((2 + \sqrt{3}) - 2) \cdot 1 = 3 - \sqrt{3} \]
\[ Q_y = 2 + (4 - 3) \cdot 1 + ((2 + \sqrt{3}) - 2) \cdot 0 = 2 + 1 = 3 \]
Thus, \(Q(3 - \sqrt{3}, 3)\).

2. **Vertex R (270° rotation)**:
- Using similar rotation formulas:
\[ R_x = C_x + (P_x - C_x) \cdot \cos(270^\circ) - (P_y - C_y) \cdot \sin(270^\circ) \]
\[ R_y = C_y + (P_x - C_x) \cdot \sin(270^\circ) + (P_y - C_y) \cdot \cos(270^\circ) \]
Substituting values:
\[ R_x = 3 + (4 - 3) \cdot 0 - ((2 + \sqrt{3}) - 2) \cdot (-1) = 3 + \sqrt{3} \]
\[ R_y = 2 + (4 - 3) \cdot (-1) + ((2 + \sqrt{3}) - 2) \cdot 0 = 2 - 1 = 1 \]
Thus, \(R(3 + \sqrt{3}, 1)\).

3. **Vertex S (180° rotation)**:
- This vertex can be calculated as:
\[ S_x = 2C_x - P_x = 2 \cdot 3 - 4 = 2 \]
\[ S_y = 2C_y - P_y = 2 \cdot 2 - (2 + \sqrt{3}) = 2 - \sqrt{3} \]
Thus, \(S(2, 2 - \sqrt{3})\).

### Final Answer
The other vertices of the square are:
- \(Q(3 - \sqrt{3}, 3)\)
- \(R(3 + \sqrt{3}, 1)\)
- \(S(2, 2 - \sqrt{3})\)