The acute angle between the lines joining the orign to points of intersection of the line `sqrt(3)x+y=2` and the circle `x^(2)+y^(2)=4` is

Options

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

Answer

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

### Step 1: Find the points of intersection
We start with the equations of the line and the circle:
1. Line: \( \sqrt{3}x + y = 2 \)
2. Circle: \( x^2 + y^2 = 4 \)

We can express \( y \) from the line equation:
\[ y = 2 - \sqrt{3}x \]

Now, substitute this expression for \( y \) into the circle equation:
\[ x^2 + (2 - \sqrt{3}x)^2 = 4 \]

### Step 2: Simplify the equation
Expanding the equation:
\[ x^2 + (2 - \sqrt{3}x)^2 = 4 \]
\[ x^2 + (4 - 4\sqrt{3}x + 3x^2) = 4 \]
Combine like terms:
\[ 4x^2 - 4\sqrt{3}x + 4 - 4 = 0 \]
This simplifies to:
\[ 4x^2 - 4\sqrt{3}x = 0 \]

### Step 3: Factor the equation
Factoring out \( 4x \):
\[ 4x(x - \sqrt{3}) = 0 \]

### Step 4: Solve for \( x \)
Setting each factor to zero gives:
1. \( 4x = 0 \) → \( x = 0 \)
2. \( x - \sqrt{3} = 0 \) → \( x = \sqrt{3} \)

### Step 5: Find corresponding \( y \) values
Substituting \( x = 0 \) into the line equation:
\[ y = 2 - \sqrt{3}(0) = 2 \]
So one point of intersection is \( (0, 2) \).

Now substituting \( x = \sqrt{3} \):
\[ y = 2 - \sqrt{3}(\sqrt{3}) = 2 - 3 = -1 \]
So the other point of intersection is \( (\sqrt{3}, -1) \).

### Step 6: Find slopes of the lines from origin to points of intersection
The slopes of the lines from the origin to the points \( (0, 2) \) and \( (\sqrt{3}, -1) \) are:
1. Slope to \( (0, 2) \): \( m_1 = \frac{2 - 0}{0 - 0} \) (undefined, vertical line)
2. Slope to \( (\sqrt{3}, -1) \): \( m_2 = \frac{-1 - 0}{\sqrt{3} - 0} = \frac{-1}{\sqrt{3}} \)

### Step 7: Use the formula for angle between two lines
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 \( m_2 \):
\[ \tan \theta = \left| \frac{0 - (-\frac{1}{\sqrt{3}})}{1 + 0} \right| = \frac{1}{\sqrt{3}} \]

### Step 8: Find the angle
Now, we find \( \theta \):
\[ \theta = \tan^{-1}\left(\frac{1}{\sqrt{3}}\right) = 30^\circ \]

### Conclusion
Thus, the acute angle between the lines joining the origin to the points of intersection is \( 30^\circ \) or \( \frac{\pi}{6} \) radians.

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