A ray of light incident at the point (-2, -1) gets reflected from the tangent at (0, -1) to the circle `x^2 +y^2=1`. The reflected ray touches the circle. The equation of the line along which the incident ray moved is

Options

A. `4x-2y +11=0`
B. ` 4x + 3y +11=0`
C. `3x + 4y +11=0`
D. None of these

Answer

To solve the problem, we need to find the equation of the line along which the incident ray moved, given that it reflects off the tangent to the circle at the point (0, -1). The equation of the circle is \(x^2 + y^2 = 1\).

### Step-by-Step Solution:

1. **Identify the Circle and Tangent Point**:
The given circle is \(x^2 + y^2 = 1\). The center of the circle is at (0, 0) and the radius is 1. The tangent point provided is (0, -1).

2. **Equation of the Tangent Line**:
The tangent line at the point (0, -1) is horizontal because it is parallel to the x-axis. Therefore, its equation is:
\[ y = -1 \]

3. **Identify the Incident Ray Point**:
The ray of light is incident at the point (-2, -1). This point lies on the tangent line \(y = -1\).

4. **Calculate the Slope of the Incident Ray**:
Let the slope of the incident ray be \(m\). The incident ray passes through the point (-2, -1). Using the point-slope form of the line, the equation of the incident ray can be written as:
\[ y + 1 = m(x + 2) \]

5. **Find the Condition for Tangency**:
The reflected ray must touch the circle. The condition for tangency of a line \(y = mx + c\) with the circle \(x^2 + y^2 = 1\) is given by:
\[ c^2 = r^2(1 + m^2) \]
Here, \(c\) is the y-intercept of the line, which can be expressed as \(c = -1 + 2m\) (from the equation of the incident ray).

6. **Substituting into the Tangency Condition**:
Substitute \(c = -1 + 2m\) into the tangency condition:
\[ (-1 + 2m)^2 = 1(1 + m^2) \]
Expanding both sides:
\[ 1 - 4m + 4m^2 = 1 + m^2 \]
Rearranging gives:
\[ 3m^2 - 4m = 0 \]
Factoring out \(m\):
\[ m(3m - 4) = 0 \]
Thus, \(m = 0\) or \(m = \frac{4}{3}\).

7. **Determine the Slope of the Reflected Ray**:
The slope of the reflected ray is the negative of the tangent slope, so:
\[ m_{\text{reflected}} = -m \]
If \(m = \frac{4}{3}\), then \(m_{\text{reflected}} = -\frac{4}{3}\).

8. **Equation of the Reflected Ray**:
Using the point-slope form for the reflected ray at the point (0, -1):
\[ y + 1 = -\frac{4}{3}(x - 0) \]
Simplifying gives:
\[ y + 1 = -\frac{4}{3}x \]
Rearranging gives:
\[ 4x + 3y + 3 = 0 \]

9. **Conclusion**:
The equation of the line along which the incident ray moved is:
\[ 4x + 3y + 3 = 0 \]