The acute angle formed between the lines joining the origin to the points of intersection of the curves `x^2+y^2-2x-1=0` and `x+y=1`, is

Options

A. `tan^(-1)(-(1)/(2))`
B. `tan^(-1)(2)`
C. `tan^(-1)((1)/(2))`
D. `60^@`

Explanation

To find the acute angle formed between the lines joining the origin to the points of intersection of the curves \( x^2 + y^2 - 2x - 1 = 0 \) and \( x + y = 1 \), we can follow these steps:

### Step 1: Identify the curves
The first curve is given by:
\[ x^2 + y^2 - 2x - 1 = 0 \]
This can be rearranged to represent a circle.

The second curve is a straight line:
\[ x + y = 1 \]

### Step 2: Rewrite the first curve
To rewrite the first curve in standard form, we complete the square:
\[ x^2 - 2x + y^2 - 1 = 0 \]
\[ (x - 1)^2 + y^2 = 2 \]
This represents a circle centered at \( (1, 0) \) with a radius of \( \sqrt{2} \).

### Step 3: Find the intersection points
Now, we will substitute \( y = 1 - x \) from the second equation into the first equation:
\[ (x - 1)^2 + (1 - x)^2 = 2 \]
Expanding this:
\[ (x - 1)^2 + (1 - x)^2 = (x - 1)^2 + (x - 1)^2 = 2(x - 1)^2 = 2 \]
\[ (x - 1)^2 = 1 \]
Taking the square root:
\[ x - 1 = \pm 1 \]
Thus, we find:
\[ x = 2 \quad \text{or} \quad x = 0 \]
Now substituting back to find \( y \):
- If \( x = 2 \): \( y = 1 - 2 = -1 \) (Point \( (2, -1) \))
- If \( x = 0 \): \( y = 1 - 0 = 1 \) (Point \( (0, 1) \))

### Step 4: Find the slopes of the lines from the origin
The points of intersection are \( (2, -1) \) and \( (0, 1) \).

1. For the point \( (2, -1) \):
- Slope \( m_1 = \frac{-1 - 0}{2 - 0} = \frac{-1}{2} \)

2. For the point \( (0, 1) \):
- Slope \( m_2 = \frac{1 - 0}{0 - 0} \) (This is a vertical line, slope is undefined)

### Step 5: Calculate the angle between the lines
To find the angle \( \theta \) between the two lines, we use the formula:
\[ \tan \theta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right| \]
Since \( m_2 \) is undefined (vertical line), we can directly find the angle between the horizontal line (slope = 0) and the line with slope \( m_1 = -\frac{1}{2} \).

Using the formula for the angle between a line and the x-axis:
\[ \tan \theta = \left| m_1 \right| = \frac{1}{2} \]
Thus:
\[ \theta = \tan^{-1} \left( \frac{1}{2} \right) \]

### Final Result
The acute angle formed between the lines joining the origin to the points of intersection is:
\[ \theta = \tan^{-1} \left( \frac{1}{2} \right) \]