The circle with equation `x^2 +y^2 = 1` intersects the line `y= 7x+5` at two distinct points A and B. Let C be the point at which the positive x-axis intersects the circle. The angle ACB is

Options

A. `tan^(-1)((4)/(3))`
B. `tan^(-1)((3)/(4))`
C. `pi//4`
D. `tan^(-1)((3)/(2))`

Explanation

To solve the problem step by step, we will follow these procedures:

### Step 1: Identify the equations
We have the equation of the circle:
\[ x^2 + y^2 = 1 \]
And the equation of the line:
\[ y = 7x + 5 \]

### Step 2: Substitute the line equation into the circle equation
To find the points of intersection A and B, we substitute \( y \) from the line equation into the circle equation:
\[ x^2 + (7x + 5)^2 = 1 \]

### Step 3: Expand and simplify
Expanding the equation:
\[ x^2 + (49x^2 + 70x + 25) = 1 \]
Combine like terms:
\[ 50x^2 + 70x + 25 - 1 = 0 \]
\[ 50x^2 + 70x + 24 = 0 \]

### Step 4: Solve the quadratic equation
Now we will use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) where \( a = 50, b = 70, c = 24 \):
- Calculate the discriminant:
\[ b^2 - 4ac = 70^2 - 4 \cdot 50 \cdot 24 = 4900 - 4800 = 100 \]
- Now apply the quadratic formula:
\[ x = \frac{-70 \pm \sqrt{100}}{2 \cdot 50} \]
\[ x = \frac{-70 \pm 10}{100} \]
This gives us two values for \( x \):
1. \( x_1 = \frac{-60}{100} = -\frac{3}{5} \)
2. \( x_2 = \frac{-80}{100} = -\frac{4}{5} \)

### Step 5: Find corresponding y-values
Now substitute \( x_1 \) and \( x_2 \) back into the line equation to find \( y \):
1. For \( x_1 = -\frac{3}{5} \):
\[ y_1 = 7(-\frac{3}{5}) + 5 = -\frac{21}{5} + 5 = -\frac{21}{5} + \frac{25}{5} = \frac{4}{5} \]
So, point A is \( A(-\frac{3}{5}, \frac{4}{5}) \).

2. For \( x_2 = -\frac{4}{5} \):
\[ y_2 = 7(-\frac{4}{5}) + 5 = -\frac{28}{5} + 5 = -\frac{28}{5} + \frac{25}{5} = -\frac{3}{5} \]
So, point B is \( B(-\frac{4}{5}, -\frac{3}{5}) \).

### Step 6: Identify point C
Point C is where the positive x-axis intersects the circle, which is at:
\[ C(1, 0) \]

### Step 7: Find slopes of lines AC and BC
- Slope of line AC:
\[ \text{slope of AC} = \frac{y_C - y_A}{x_C - x_A} = \frac{0 - \frac{4}{5}}{1 - (-\frac{3}{5})} = \frac{-\frac{4}{5}}{1 + \frac{3}{5}} = \frac{-\frac{4}{5}}{\frac{8}{5}} = -\frac{4}{8} = -\frac{1}{2} \]

- Slope of line BC:
\[ \text{slope of BC} = \frac{y_C - y_B}{x_C - x_B} = \frac{0 - (-\frac{3}{5})}{1 - (-\frac{4}{5})} = \frac{\frac{3}{5}}{1 + \frac{4}{5}} = \frac{\frac{3}{5}}{\frac{9}{5}} = \frac{3}{9} = \frac{1}{3} \]

### Step 8: Calculate angle ACB
Using the formula for the angle between two lines:
\[ \tan(\theta) = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right| \]
Where \( m_1 = -\frac{1}{2} \) and \( m_2 = \frac{1}{3} \):
\[ \tan(\theta) = \left| \frac{-\frac{1}{2} - \frac{1}{3}}{1 + (-\frac{1}{2})(\frac{1}{3})} \right| \]
Calculating the numerator:
\[ -\frac{1}{2} - \frac{1}{3} = -\frac{3}{6} - \frac{2}{6} = -\frac{5}{6} \]
Calculating the denominator:
\[ 1 - \frac{1}{6} = \frac{5}{6} \]
Thus:
\[ \tan(\theta) = \left| \frac{-\frac{5}{6}}{\frac{5}{6}} \right| = 1 \]
This means:
\[ \theta = 45^\circ \]

### Final Answer
The angle \( ACB \) is \( 45^\circ \).

---