Two fixed charges -2Q and Q are located at the point with coordinates (-3a,0) and (+3a,0) respectively in the x- y plane .
Show that all points in the x-y plane where the electric potential due to the two charge is zero , lie on a circle. If its radius is r = xa and the location of its centre is (ya,0) find x and y .

Explanation

To solve the problem, we need to find the points in the x-y plane where the electric potential due to the two charges is zero and show that these points lie on a circle. We will also determine the values of \( x \) and \( y \) based on the given conditions.

### Step-by-Step Solution:

1. **Identify the Charges and Their Positions**:
- We have two charges: \( -2Q \) located at \( (-3a, 0) \) and \( Q \) located at \( (3a, 0) \).

2. **Write the Expression for Electric Potential**:
- The electric potential \( V \) at a point \( (x, y) \) due to a point charge is given by:
\[ V = k \frac{Q}{r} \]
- Therefore, the total potential \( V \) at point \( (x, y) \) due to both charges is:
\[ V = V_1 + V_2 \]
- Where:
\[ V_1 = k \frac{-2Q}{R_1} \quad \text{and} \quad V_2 = k \frac{Q}{R_2} \]
- Here, \( R_1 \) is the distance from \( (x, y) \) to \( (-3a, 0) \) and \( R_2 \) is the distance from \( (x, y) \) to \( (3a, 0) \).

3. **Calculate the Distances \( R_1 \) and \( R_2 \)**:
- The distances are given by:
\[ R_1 = \sqrt{(x + 3a)^2 + y^2} \]
\[ R_2 = \sqrt{(x - 3a)^2 + y^2} \]

4. **Set Up the Equation for Zero Potential**:
- To find where the total potential is zero, we set:
\[ V_1 + V_2 = 0 \]
- This leads to:
\[ k \frac{-2Q}{R_1} + k \frac{Q}{R_2} = 0 \]
- Simplifying gives:
\[ -2 \frac{1}{R_1} + \frac{1}{R_2} = 0 \]
- Rearranging results in:
\[ 2R_2 = R_1 \]

5. **Substituting the Distances**:
- Substitute \( R_1 \) and \( R_2 \) into the equation:
\[ 2\sqrt{(x - 3a)^2 + y^2} = \sqrt{(x + 3a)^2 + y^2} \]

6. **Squaring Both Sides**:
- Squaring both sides to eliminate the square roots:
\[ 4((x - 3a)^2 + y^2) = (x + 3a)^2 + y^2 \]

7. **Expanding and Simplifying**:
- Expanding gives:
\[ 4(x^2 - 6ax + 9a^2 + y^2) = x^2 + 6ax + 9a^2 + y^2 \]
- Simplifying leads to:
\[ 4x^2 - 24ax + 36a^2 + 4y^2 = x^2 + 6ax + 9a^2 + y^2 \]
- Rearranging gives:
\[ 3x^2 - 30ax + 27a^2 + 3y^2 = 0 \]

8. **Dividing by 3**:
- Dividing the entire equation by 3:
\[ x^2 - 10ax + 9a^2 + y^2 = 0 \]

9. **Completing the Square**:
- Completing the square for \( x \):
\[ (x - 5a)^2 + y^2 = 16a^2 \]

10. **Identifying the Circle**:
- This is the equation of a circle with center at \( (5a, 0) \) and radius \( 4a \).

11. **Finding \( x \) and \( y \)**:
- Given \( r = xa \) and the center at \( (ya, 0) \):
- From the radius, \( r = 4a \) implies \( x = 4 \).
- The center \( (5a, 0) \) gives \( y = 5 \).

### Final Answer:
- \( x = 4 \)
- \( y = 5 \)