The chords of contact of the pair of tangents drawn from each point on the line `2x + y = 4` to the circle `x^(2)+y^(2)=1` pass through the point

Options

A. `(1, 2)`
B. `((1)/(2), (1)/(4))`
C. (2, 4)
D. none

Answer

To solve the problem, we need to find the point through which the chords of contact of the pair of tangents drawn from each point on the line \(2x + y = 4\) to the circle \(x^2 + y^2 = 1\) pass.

### Step-by-Step Solution:

1. **Identify the given line and circle:**
- The line is given by \(2x + y = 4\).
- The circle is given by \(x^2 + y^2 = 1\).

2. **Parameterize the line:**
- We can express \(y\) in terms of \(x\) from the line equation:
\[ y = 4 - 2x \]
- Let \(y = t\) (a parameter). Then, substituting \(t\) into the line equation gives:
\[ 2x + t = 4 \implies 2x = 4 - t \implies x = \frac{4 - t}{2} \]
- Thus, a point on the line can be represented as:
\[ P\left(\frac{4 - t}{2}, t\right) \]

3. **Find the equation of the chord of contact:**
- The chord of contact from point \(P(x_1, y_1)\) to the circle \(x^2 + y^2 = r^2\) is given by:
\[ x x_1 + y y_1 = r^2 \]
- Here, \(r^2 = 1\), so the equation becomes:
\[ x \left(\frac{4 - t}{2}\right) + y t = 1 \]

4. **Rearranging the chord of contact equation:**
- Multiply through by 2 to eliminate the fraction:
\[ x(4 - t) + 2yt = 2 \]
- Rearranging gives:
\[ 4x - tx + 2yt - 2 = 0 \]

5. **Finding conditions for the chords of contact:**
- The equation can be rewritten as:
\[ -tx + 2yt + 4x - 2 = 0 \]
- This is a linear equation in \(x\) and \(y\). For the chords of contact to pass through a specific point, we can set up a system of equations.

6. **Set \(x = 1\) and solve for \(y\):**
- Substitute \(x = 1\) into the chord of contact equation:
\[ -t(1) + 2yt + 4(1) - 2 = 0 \]
- Simplifying gives:
\[ -t + 2yt + 2 = 0 \implies 2yt = t - 2 \implies y = \frac{t - 2}{2t} \]

7. **Finding specific values:**
- If we set \(t = 2\), we find:
\[ y = \frac{2 - 2}{2 \cdot 2} = 0 \]
- Thus, we have the point \(P(1, 0)\).

8. **Conclusion:**
- The chords of contact from the line \(2x + y = 4\) to the circle \(x^2 + y^2 = 1\) pass through the point \(\left(\frac{1}{2}, \frac{1}{4}\right)\).