Let the equation of the circle is `x^(2) + y^(2) - 2x-4y + 1 = 0` A line through `P(alpha, -1)` is drawn which intersect the given circle at the point A and B. if PA PB has the minimum value then the value of `alpha` is.

Answer

To solve the problem, we need to find the value of \( \alpha \) such that the product \( PA \cdot PB \) is minimized, where \( P(\alpha, -1) \) is a point from which a line intersects the circle at points \( A \) and \( B \).

### Step-by-Step Solution:

1. **Rewrite the Circle Equation**:
The given equation of the circle is:
\[ x^2 + y^2 - 2x - 4y + 1 = 0 \]
We can rearrange it to find the center and radius of the circle.

2. **Complete the Square**:
For the \( x \) terms:
\[ x^2 - 2x = (x - 1)^2 - 1 \]
For the \( y \) terms:
\[ y^2 - 4y = (y - 2)^2 - 4 \]
Substituting these into the circle equation gives:
\[ (x - 1)^2 - 1 + (y - 2)^2 - 4 + 1 = 0 \]
Simplifying this, we have:
\[ (x - 1)^2 + (y - 2)^2 = 4 \]
Thus, the center of the circle is \( (1, 2) \) and the radius \( r = 2 \).

3. **Find the Distance from Point \( P \) to the Center**:
The distance \( PT \) from point \( P(\alpha, -1) \) to the center \( C(1, 2) \) is calculated using the distance formula:
\[ PT = \sqrt{(\alpha - 1)^2 + (-1 - 2)^2} = \sqrt{(\alpha - 1)^2 + 9} \]

4. **Use the Tangent-Secant Theorem**:
According to the tangent-secant theorem, we have:
\[ PA \cdot PB = PT^2 \]
Therefore:
\[ PA \cdot PB = (\sqrt{(\alpha - 1)^2 + 9})^2 = (\alpha - 1)^2 + 9 \]

5. **Minimize the Expression**:
We need to minimize:
\[ f(\alpha) = (\alpha - 1)^2 + 9 \]
The minimum value of \( f(\alpha) \) occurs when \( (\alpha - 1)^2 \) is minimized, which is at \( \alpha = 1 \).

6. **Verify the Minimum**:
To confirm that this is a minimum, we can take the second derivative:
\[ f'(\alpha) = 2(\alpha - 1) \]
Setting \( f'(\alpha) = 0 \) gives \( \alpha = 1 \). The second derivative:
\[ f''(\alpha) = 2 > 0 \]
indicates that \( f(\alpha) \) is concave up, confirming a minimum at \( \alpha = 1 \).

### Final Answer:
The value of \( \alpha \) for which \( PA \cdot PB \) has the minimum value is:
\[ \alpha = 1 \]