Consider points `A(sqrt13,0) and B(2sqrt13,0)` lying on x-axis. These points are rotated anticlockwise direction about the origin through an angle of `tan^-1(2/3)`. Let the new position of A and B be A' and B' respectively. With A' as centre and radius `2sqrt13/3` a circle `C_1` is drawn and with B' as centre and radius `sqrt13/3` circle `C_2`, is drawn. The radical axis of `C_1 and C_2` is (a) `3x+2y=20` (b) `3x+2y=10` (c) `9x+6y=65` (d) none of these

Options

A. `3x+2y=20`
B. `3x+2y=10`
C. `9x+6y=65`
D. none of these

Explanation

To solve the problem step by step, we will follow the outlined procedure based on the information provided in the question.

### Step 1: Determine the coordinates of A' and B'
Given the points A(√13, 0) and B(2√13, 0), we need to rotate these points anticlockwise about the origin through an angle of θ = tan⁻¹(2/3).

Using the rotation formulas:
- For point A:
- \( A' = (x_A \cos \theta - y_A \sin \theta, x_A \sin \theta + y_A \cos \theta) \)
- Since \( A = (√13, 0) \), we have:
- \( A' = (√13 \cdot \cos \theta, √13 \cdot \sin \theta) \)

- For point B:
- \( B' = (x_B \cos \theta - y_B \sin \theta, x_B \sin \theta + y_B \cos \theta) \)
- Since \( B = (2√13, 0) \), we have:
- \( B' = (2√13 \cdot \cos \theta, 2√13 \cdot \sin \theta) \)

### Step 2: Calculate cos(θ) and sin(θ)
From the given \( \tan \theta = \frac{2}{3} \):
- We can find \( \sin \theta \) and \( \cos \theta \) using the identity:
- \( \sin^2 \theta + \cos^2 \theta = 1 \)

Using the triangle with opposite side = 2 and adjacent side = 3:
- Hypotenuse \( h = \sqrt{2^2 + 3^2} = \sqrt{13} \)
- Thus, \( \sin \theta = \frac{2}{\sqrt{13}} \) and \( \cos \theta = \frac{3}{\sqrt{13}} \)

### Step 3: Substitute to find A' and B'
Now substituting the values of \( \sin \theta \) and \( \cos \theta \):
- For A':
- \( A' = (√13 \cdot \frac{3}{\sqrt{13}}, √13 \cdot \frac{2}{\sqrt{13}}) = (3, 2) \)

- For B':
- \( B' = (2√13 \cdot \frac{3}{\sqrt{13}}, 2√13 \cdot \frac{2}{\sqrt{13}}) = (6, 4) \)

### Step 4: Write the equations of circles C1 and C2
- Circle C1 with center A' (3, 2) and radius \( \frac{2\sqrt{13}}{3} \):
\[ (x - 3)^2 + (y - 2)^2 = \left(\frac{2\sqrt{13}}{3}\right)^2 = \frac{4 \cdot 13}{9} = \frac{52}{9} \]

- Circle C2 with center B' (6, 4) and radius \( \frac{\sqrt{13}}{3} \):
\[ (x - 6)^2 + (y - 4)^2 = \left(\frac{\sqrt{13}}{3}\right)^2 = \frac{13}{9} \]

### Step 5: Find the radical axis of C1 and C2
The radical axis is given by the equation:
\[ C1 - C2 = 0 \]

Expanding both circle equations:
1. For C1:
\[ (x - 3)^2 + (y - 2)^2 - \frac{52}{9} = 0 \]
Expanding gives:
\[ x^2 - 6x + 9 + y^2 - 4y + 4 - \frac{52}{9} = 0 \]
\[ x^2 + y^2 - 6x - 4y + \left(13 - \frac{52}{9}\right) = 0 \]
\[ x^2 + y^2 - 6x - 4y - \frac{13}{9} = 0 \]

2. For C2:
\[ (x - 6)^2 + (y - 4)^2 - \frac{13}{9} = 0 \]
Expanding gives:
\[ x^2 - 12x + 36 + y^2 - 8y + 16 - \frac{13}{9} = 0 \]
\[ x^2 + y^2 - 12x - 8y + \left(52 - \frac{13}{9}\right) = 0 \]
\[ x^2 + y^2 - 12x - 8y - \frac{413}{9} = 0 \]

### Step 6: Subtract the two equations
Subtracting the second equation from the first gives:
\[ (-6x + 12x) + (-4y + 8y) + \left(-\frac{13}{9} + \frac{413}{9}\right) = 0 \]
This simplifies to:
\[ 6x + 4y - \frac{400}{9} = 0 \]

Multiplying through by 9 to eliminate the fraction gives:
\[ 54x + 36y - 400 = 0 \]

Dividing through by 6 simplifies to:
\[ 9x + 6y - \frac{400}{6} = 0 \]

### Step 7: Final equation of the radical axis
The equation of the radical axis is:
\[ 9x + 6y = 65 \]

### Conclusion
Thus, the correct answer is option (c) \( 9x + 6y = 65 \).