Any circle through the point of intersection of the lines `x+sqrt(3)y=1` and `sqrt(3)x-y=2` intersects there lines at points `Pa n dQ` . Then the angle subtended by the are `P Q` at its center is `180^0` (b) `90^0` (c) `120^0` depends on center and radius

Options

A. `180^(@)`
B. `90^(@)`
C. `120^(@)`
D. Depends on centre and radius

Explanation

To solve the problem, we need to find the angle subtended by the arc PQ at the center of the circle, given that the circle passes through the intersection of the two lines. Let's go through the steps systematically.

### Step 1: Find the point of intersection of the lines
We have two lines given by the equations:
1. \( x + \sqrt{3}y = 1 \) (Line 1)
2. \( \sqrt{3}x - y = 2 \) (Line 2)

To find the intersection, we can solve these equations simultaneously.

**Substituting for \( x \) from Line 1 into Line 2:**
From Line 1, we can express \( x \) as:
\[ x = 1 - \sqrt{3}y \]

Now substitute this into Line 2:
\[ \sqrt{3}(1 - \sqrt{3}y) - y = 2 \]
\[ \sqrt{3} - 3y - y = 2 \]
\[ \sqrt{3} - 4y = 2 \]
\[ 4y = \sqrt{3} - 2 \]
\[ y = \frac{\sqrt{3} - 2}{4} \]

Now substituting \( y \) back into Line 1 to find \( x \):
\[ x + \sqrt{3}\left(\frac{\sqrt{3} - 2}{4}\right) = 1 \]
\[ x + \frac{3 - 2\sqrt{3}}{4} = 1 \]
\[ x = 1 - \frac{3 - 2\sqrt{3}}{4} \]
\[ x = \frac{4 - (3 - 2\sqrt{3})}{4} \]
\[ x = \frac{1 + 2\sqrt{3}}{4} \]

Thus, the point of intersection \( O \) is:
\[ O\left(\frac{1 + 2\sqrt{3}}{4}, \frac{\sqrt{3} - 2}{4}\right) \]

### Step 2: Find the slopes of the lines
Next, we need to find the slopes of both lines to determine the angle between them.

For Line 1:
\[ x + \sqrt{3}y = 1 \]
Rearranging gives:
\[ \sqrt{3}y = -x + 1 \]
\[ y = -\frac{1}{\sqrt{3}}x + \frac{1}{\sqrt{3}} \]
Thus, the slope \( m_1 = -\frac{1}{\sqrt{3}} \).

For Line 2:
\[ \sqrt{3}x - y = 2 \]
Rearranging gives:
\[ y = \sqrt{3}x - 2 \]
Thus, the slope \( m_2 = \sqrt{3} \).

### Step 3: Calculate the angle between the lines
The angle \( \theta \) between two lines with slopes \( m_1 \) and \( m_2 \) can be found using the formula:
\[ \tan(\theta) = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right| \]

Calculating \( m_1 m_2 \):
\[ m_1 m_2 = -\frac{1}{\sqrt{3}} \cdot \sqrt{3} = -1 \]

Thus:
\[ \tan(\theta) = \left| \frac{-\frac{1}{\sqrt{3}} - \sqrt{3}}{1 - 1} \right| \]
Since \( m_1 m_2 = -1 \), the lines are perpendicular, which means:
\[ \theta = 90^\circ \]

### Step 4: Find the angle subtended at the center
According to the property of circles, the angle subtended at the center \( \theta \) is double the angle subtended at the circumference:
\[ \text{Angle subtended at center} = 2 \times \theta = 2 \times 90^\circ = 180^\circ \]

### Final Answer
The angle subtended by the arc PQ at its center is:
\[ \boxed{180^\circ} \]