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 `"___________"`.

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

2. **Write the equation of the chord of contact**:
- The equation of the chord of contact from a point \((h, k)\) to the circle \(x^2 + y^2 = r^2\) is given by:
\[ hx + ky = r^2 \]
- For our circle, \(r^2 = 1\), so the equation becomes:
\[ hx + ky = 1 \]

3. **Substitute the line equation into the chord of contact**:
- From the line \(2x + y = 4\), we can express \(y\) in terms of \(x\):
\[ y = 4 - 2x \]
- Now, substitute \(y\) into the chord of contact equation:
\[ hx + k(4 - 2x) = 1 \]
- Expanding this gives:
\[ hx + 4k - 2kx = 1 \]
- Rearranging terms:
\[ (h - 2k)x + 4k = 1 \]

4. **Set up the system of equations**:
- For the equation to represent a family of lines, the coefficients of \(x\) and the constant term must be equal to zero. Thus, we have:
\[ h - 2k = 0 \quad \text{(1)} \]
\[ 4k - 1 = 0 \quad \text{(2)} \]

5. **Solve the equations**:
- From equation (2):
\[ 4k = 1 \implies k = \frac{1}{4} \]
- Substitute \(k\) back into equation (1):
\[ h - 2\left(\frac{1}{4}\right) = 0 \implies h - \frac{1}{2} = 0 \implies h = \frac{1}{2} \]

6. **Find the point (h, k)**:
- Thus, the point through which the chords of contact pass is:
\[ \left(h, k\right) = \left(\frac{1}{2}, \frac{1}{4}\right) \]

### Final Answer:
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 \(\left(\frac{1}{2}, \frac{1}{4}\right)\).