Tangents PA and PB to the director circle of circle `x^(2)+y^(2)-2x-4y-5=0` from `P(1+3sqrt(3), 2+3sqrt(3))` include an angle `angleAPB = alpha`, then `cos alpha` is :

Options

A. Positive
B. negative
C. Zero
D. Data insufficient

Explanation

To solve the problem, we need to find the cosine of the angle \( \alpha \) formed by the tangents \( PA \) and \( PB \) from the point \( P(1 + 3\sqrt{3}, 2 + 3\sqrt{3}) \) to the director circle of the given circle \( x^2 + y^2 - 2x - 4y - 5 = 0 \).

### Step 1: Find the center and radius of the given circle
The equation of the circle can be rewritten in standard form. We complete the square for both \( x \) and \( y \).

1. Rearranging:
\[ x^2 - 2x + y^2 - 4y = 5 \]

2. Completing the square:
\[ (x - 1)^2 - 1 + (y - 2)^2 - 4 = 5 \]
\[ (x - 1)^2 + (y - 2)^2 = 10 \]

Thus, the center of the circle is \( (1, 2) \) and the radius \( r = \sqrt{10} \).

### Step 2: Find the equation of the director circle
The director circle of a circle with center \( (h, k) \) and radius \( r \) is given by:
\[ (x - h)^2 + (y - k)^2 = 2r^2 \]

Substituting \( h = 1 \), \( k = 2 \), and \( r^2 = 10 \):
\[ (x - 1)^2 + (y - 2)^2 = 20 \]

### Step 3: Find the distance from point \( P \) to the center of the director circle
The distance \( d \) from point \( P(1 + 3\sqrt{3}, 2 + 3\sqrt{3}) \) to the center \( (1, 2) \) is calculated as:
\[ d = \sqrt{(1 + 3\sqrt{3} - 1)^2 + (2 + 3\sqrt{3} - 2)^2} \]
\[ = \sqrt{(3\sqrt{3})^2 + (3\sqrt{3})^2} = \sqrt{27 + 27} = \sqrt{54} = 3\sqrt{6} \]

### Step 4: Use the formula for cosines of the angle between tangents
The cosine of the angle \( \alpha \) between the two tangents from point \( P \) to the director circle is given by:
\[ \cos \alpha = \frac{d^2 - r^2}{d^2 + r^2} \]

Substituting \( d = 3\sqrt{6} \) and \( r = \sqrt{20} \):
1. Calculate \( d^2 \):
\[ d^2 = (3\sqrt{6})^2 = 54 \]

2. Calculate \( r^2 \):
\[ r^2 = 20 \]

3. Substitute into the cosine formula:
\[ \cos \alpha = \frac{54 - 20}{54 + 20} = \frac{34}{74} = \frac{17}{37} \]

### Final Answer
Thus, the value of \( \cos \alpha \) is:
\[ \cos \alpha = \frac{17}{37} \]