Consider the family of circles `x^(2)+y^(2)-2x-2ay-8=0` passing through two fixed points A and B . Also, `S=0` is a cricle of this family, the tangent to which at A and B intersect on the line `x+2y+5=0`.
The distance between the points A and B , is

Options

A. 4
B. `4 sqrt(2)`
C. 6
D. 8

Explanation

To solve the problem, we need to find the distance between the two fixed points A and B that lie on the family of circles given by the equation:

\[ x^2 + y^2 - 2x - 2ay - 8 = 0 \]

### Step-by-Step Solution:

1. **Identify the Circle Equation**:
The equation of the circle can be rewritten as:
\[ x^2 + y^2 - 2x - 2ay - 8 = 0 \]
This represents a family of circles parameterized by \( a \).

2. **Set \( S = 0 \)**:
We are given that \( S = 0 \) is a circle of this family. Thus, we set:
\[ S = x^2 + y^2 - 2x - 8 = 0 \]
This simplifies to:
\[ x^2 + y^2 - 2x - 8 = 0 \]

3. **Find the Tangent Condition**:
The tangents at points A and B intersect on the line:
\[ x + 2y + 5 = 0 \]

4. **Substituting \( y = 0 \)**:
To find the points A and B, we can substitute \( y = 0 \) into the circle equation:
\[ x^2 - 2x - 8 = 0 \]

5. **Solve the Quadratic Equation**:
The quadratic equation can be solved using the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \( a = 1, b = -2, c = -8 \):
\[ x = \frac{2 \pm \sqrt{(-2)^2 - 4 \cdot 1 \cdot (-8)}}{2 \cdot 1} = \frac{2 \pm \sqrt{4 + 32}}{2} = \frac{2 \pm \sqrt{36}}{2} = \frac{2 \pm 6}{2} \]
This gives us:
\[ x = \frac{8}{2} = 4 \quad \text{and} \quad x = \frac{-4}{2} = -2 \]

6. **Coordinates of Points A and B**:
Therefore, the coordinates of points A and B are:
\[ A(4, 0) \quad \text{and} \quad B(-2, 0) \]

7. **Calculate the Distance Between A and B**:
The distance \( AB \) can be calculated using the distance formula:
\[ AB = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]
Substituting the coordinates of A and B:
\[ AB = \sqrt{(-2 - 4)^2 + (0 - 0)^2} = \sqrt{(-6)^2} = \sqrt{36} = 6 \]

### Conclusion:
The distance between points A and B is \( 6 \) units.