If the line `xcostheta=2` is the equation of a transverse common tangent to the circles `x^2+y^2=4` and `x^2+y^2-6sqrt(3)x-6y+20=0` , then the value of `theta` is `(5pi)/6` (b) `(2pi)/3` (c) `pi/3` (d) `pi/6`

Options

A. `5pi //6`
B. `2pi //3`
C. `pi //3`
D. `pi//6`

Explanation

To solve the problem, we need to find the value of \( \theta \) such that the line defined by the equation \( x \cos \theta = 2 \) is a transverse common tangent to the given circles.

### Step 1: Identify the circles
The first circle is given by the equation:
\[ x^2 + y^2 = 4 \]
This represents a circle centered at \( (0, 0) \) with a radius \( r_1 = 2 \).

The second circle is given by the equation:
\[ x^2 + y^2 - 6\sqrt{3}x - 6y + 20 = 0 \]
We can rewrite this in standard form. Completing the square for \( x \) and \( y \):
\[ (x^2 - 6\sqrt{3}x) + (y^2 - 6y) + 20 = 0 \]
\[ (x - 3\sqrt{3})^2 - 27 + (y - 3)^2 - 9 + 20 = 0 \]
\[ (x - 3\sqrt{3})^2 + (y - 3)^2 = 16 \]
This represents a circle centered at \( (3\sqrt{3}, 3) \) with a radius \( r_2 = 4 \).

### Step 2: Calculate the distance between the centers
The distance \( d \) between the centers of the two circles \( C_1(0, 0) \) and \( C_2(3\sqrt{3}, 3) \) is given by:
\[ d = \sqrt{(3\sqrt{3} - 0)^2 + (3 - 0)^2} = \sqrt{27 + 9} = \sqrt{36} = 6 \]

### Step 3: Check if the circles touch externally
The circles touch externally if:
\[ d = r_1 + r_2 \]
Calculating \( r_1 + r_2 \):
\[ r_1 + r_2 = 2 + 4 = 6 \]
Since \( d = 6 \), the circles do indeed touch externally.

### Step 4: Find the equation of the common tangent
The equation of the common tangent can be derived from the centers and the radii. The equation of the common tangent at the point of tangency can be expressed as:
\[ \sqrt{3}x + y - 4 = 0 \]

### Step 5: Compare with the given line equation
The given line equation is:
\[ x \cos \theta = 2 \]
This can be rewritten as:
\[ x \cos \theta + y \sin \theta = 2 \]
To find \( \theta \), we compare the coefficients from the common tangent equation \( \sqrt{3}x + y - 4 = 0 \) with the form \( r \cos \theta + y \sin \theta = 2 \).

From the common tangent equation:
- Coefficient of \( x \) is \( \sqrt{3} \)
- Coefficient of \( y \) is \( 1 \)

### Step 6: Find \( \theta \)
From the comparison:
\[ r \cos \theta = \sqrt{3} \quad \text{and} \quad r \sin \theta = 1 \]
Using the identity \( r = 2 \):
\[ 2 \cos \theta = \sqrt{3} \quad \Rightarrow \quad \cos \theta = \frac{\sqrt{3}}{2} \]
\[ 2 \sin \theta = 1 \quad \Rightarrow \quad \sin \theta = \frac{1}{2} \]
The angle \( \theta \) that satisfies both conditions is:
\[ \theta = \frac{\pi}{6} \]

### Final Answer
Thus, the value of \( \theta \) is:
\[ \theta = \frac{\pi}{6} \]