The acute angle between lines joining origin and intersection points of line `sqrt3x + y - 2 = 0` and the circle `x^2 + y^2 = 4` is

Options

A. `(pi)/(6)`
B. `(pi)/(4)`
C. `(pi)/(3)`
D. `(pi)/(2)`

Explanation

To find the acute angle between the lines joining the origin and the intersection points of the line \( \sqrt{3}x + y - 2 = 0 \) and the circle \( x^2 + y^2 = 4 \), we can follow these steps:

### Step 1: Find the intersection points of the line and the circle

We start by substituting the equation of the line into the equation of the circle.

1. Rearranging the line equation:
\[ y = 2 - \sqrt{3}x \]

2. Substitute \( y \) in the circle's equation:
\[ x^2 + (2 - \sqrt{3}x)^2 = 4 \]

3. Expanding the equation:
\[ x^2 + (4 - 4\sqrt{3}x + 3x^2) = 4 \]
\[ 4x^2 - 4\sqrt{3}x + 4 - 4 = 0 \]
\[ 4x^2 - 4\sqrt{3}x = 0 \]

4. Factoring out \( 4x \):
\[ 4x(x - \sqrt{3}) = 0 \]

5. This gives us \( x = 0 \) or \( x = \sqrt{3} \).

### Step 2: Find corresponding \( y \) values

1. For \( x = 0 \):
\[ y = 2 - \sqrt{3}(0) = 2 \quad \Rightarrow \quad (0, 2) \]

2. For \( x = \sqrt{3} \):
\[ y = 2 - \sqrt{3}(\sqrt{3}) = 2 - 3 = -1 \quad \Rightarrow \quad (\sqrt{3}, -1) \]

### Step 3: Find the slopes of the lines from the origin to the intersection points

1. Slope to point \( (0, 2) \):
\[ m_1 = \frac{2 - 0}{0 - 0} \quad \text{(undefined, vertical line)} \]

2. Slope to point \( (\sqrt{3}, -1) \):
\[ m_2 = \frac{-1 - 0}{\sqrt{3} - 0} = \frac{-1}{\sqrt{3}} = -\frac{1}{\sqrt{3}} \]

### Step 4: Use the formula for the angle between two lines

The formula for the angle \( \theta \) between two lines with slopes \( m_1 \) and \( m_2 \) is given by:
\[ \tan \theta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right| \]

Since \( m_1 \) is undefined (vertical line), we can directly find the angle between the vertical line and the line with slope \( -\frac{1}{\sqrt{3}} \).

### Step 5: Calculate the angle

1. The angle \( \theta \) between the vertical line and the line with slope \( -\frac{1}{\sqrt{3}} \) is:
\[ \tan \theta = \left| -\frac{1}{\sqrt{3}} \right| = \frac{1}{\sqrt{3}} \]

2. Therefore, \( \theta = 30^\circ \).

### Step 6: Find the acute angle between the two lines

Since one line is vertical and the other has a slope of \( -\frac{1}{\sqrt{3}} \), the acute angle between them is:
\[ 90^\circ - 30^\circ = 60^\circ \]

### Final Answer

The acute angle between the lines joining the origin and the intersection points is \( 60^\circ \).

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