The exhaustive range of value of a such that the angle between the pair of tangents drawn from `(a,a)` to the circle `x^2 +y^2 -2x -2y -6 =0` lies in the range `(pi/3,pi)` is

Options

A. (-1,3)
B. `(-5,-3)uu(3,5)`
C. `(-3,5)`
D. `(-3,-1)uu(3,5)`

Answer

To solve the problem, we need to find the exhaustive range of values of \( a \) such that the angle between the pair of tangents drawn from the point \( (a, a) \) to the circle given by the equation \( x^2 + y^2 - 2x - 2y - 6 = 0 \) lies in the range \( \left( \frac{\pi}{3}, \pi \right) \).

### Step 1: Rewrite the equation of the circle
The given equation of the circle is:
\[ x^2 + y^2 - 2x - 2y - 6 = 0 \]
We can rewrite it in standard form by completing the square:
\[ (x^2 - 2x) + (y^2 - 2y) = 6 \]
Completing the square for \( x \) and \( y \):
\[ (x - 1)^2 - 1 + (y - 1)^2 - 1 = 6 \]
\[ (x - 1)^2 + (y - 1)^2 = 8 \]
This shows that the center of the circle is \( (1, 1) \) and the radius \( R \) is \( \sqrt{8} = 2\sqrt{2} \).

### Step 2: Determine the angle between the tangents
The angle \( 2\alpha \) between the pair of tangents from point \( (a, a) \) to the circle can be expressed in terms of the distance from the point to the center of the circle and the radius of the circle. The angle \( 2\alpha \) lies between \( \frac{\pi}{3} \) and \( \pi \).

### Step 3: Use the tangent angle formula
The condition for the angle \( 2\alpha \) is given by:
\[ \sin \alpha < \frac{R}{CP} \]
where \( CP \) is the distance from the point \( (a, a) \) to the center \( (1, 1) \).

### Step 4: Calculate the distance \( CP \)
The distance \( CP \) is calculated as:
\[ CP = \sqrt{(a - 1)^2 + (a - 1)^2} = \sqrt{2(a - 1)^2} = \sqrt{2} |a - 1| \]

### Step 5: Set up the inequalities
From the angle condition, we have:
\[ \sin \alpha < \frac{R}{CP} \implies \sin \alpha < \frac{2\sqrt{2}}{\sqrt{2}|a - 1|} \implies \sin \alpha < \frac{2}{|a - 1|} \]
Given that \( 2\alpha \) lies between \( \frac{\pi}{3} \) and \( \pi \), we have:
\[ \frac{\sqrt{3}}{2} < \frac{2}{|a - 1|} \implies |a - 1| < \frac{4}{\sqrt{3}} \]
This gives us two inequalities:
\[ - \frac{4}{\sqrt{3}} < a - 1 < \frac{4}{\sqrt{3}} \]
Thus,
\[ 1 - \frac{4}{\sqrt{3}} < a < 1 + \frac{4}{\sqrt{3}} \]

### Step 6: Solve the inequalities
Calculating the bounds:
\[ 1 - \frac{4}{\sqrt{3}} \approx 1 - 2.309 = -1.309 \quad \text{and} \quad 1 + \frac{4}{\sqrt{3}} \approx 1 + 2.309 = 3.309 \]
So we have:
\[ -1.309 < a < 3.309 \]

### Step 7: Combine with the angle condition
Now, we also need to consider the case where \( 2\alpha \) is greater than \( \frac{\pi}{3} \):
This leads to \( CP > 2R \), which gives:
\[ \sqrt{2}|a - 1| > 4\sqrt{2} \implies |a - 1| > 4 \]
This results in two more inequalities:
\[ a - 1 < -4 \quad \text{or} \quad a - 1 > 4 \]
Thus:
\[ a < -3 \quad \text{or} \quad a > 5 \]

### Final Step: Combine all conditions
The exhaustive range of values for \( a \) is:
\[ (-\infty, -3) \cup (5, \infty) \]