Two fixed charges `-2Q` and `+Q` are located at points `(-3a,0)` and `(+3a,0)` respectively. Then which of the following statement is correct?

Options

A. Points where the electric potential due to the two charges is zero in x-y plane, lie on a circle radius 4a and centre (5a,0)
B. potential is zero at `x=a` and `x=9a`
C. both a and b are wrong
D. both a and b are correct

Explanation

To solve the problem, we need to analyze the electric potential created by two fixed charges, \(-2Q\) and \(+Q\), located at points \((-3a, 0)\) and \((3a, 0)\) respectively.

### Step-by-Step Solution:

1. **Identify the Charges and Their Positions**:
- We have two charges:
- Charge \(Q_1 = -2Q\) at position \((-3a, 0)\)
- Charge \(Q_2 = +Q\) at position \((3a, 0)\)

2. **Understanding Electric Potential**:
- The electric potential \(V\) at a point due to a point charge is given by the formula:
\[ V = k \frac{Q}{r} \]
where \(k\) is Coulomb's constant, \(Q\) is the charge, and \(r\) is the distance from the charge to the point.

3. **Setting Up the Potential Equation**:
- Let’s consider a point \(P(x, y)\) in the xy-plane where we want to find the potential. The distances from point \(P\) to the charges are:
- \(r_1 = \sqrt{(x + 3a)^2 + y^2}\) (distance to charge \(-2Q\))
- \(r_2 = \sqrt{(x - 3a)^2 + y^2}\) (distance to charge \(+Q\))
- The total potential at point \(P\) is:
\[ V = k \left( \frac{-2Q}{r_1} + \frac{Q}{r_2} \right) \]

4. **Setting the Potential to Zero**:
- We want to find the points where the total potential \(V = 0\):
\[ \frac{-2Q}{r_1} + \frac{Q}{r_2} = 0 \]
- Rearranging gives:
\[ 2 \cdot \frac{1}{r_1} = \frac{1}{r_2} \quad \Rightarrow \quad 2r_2 = r_1 \]

5. **Substituting the Distances**:
- Substitute \(r_1\) and \(r_2\):
\[ 2\sqrt{(x - 3a)^2 + y^2} = \sqrt{(x + 3a)^2 + y^2} \]
- Squaring both sides:
\[ 4((x - 3a)^2 + y^2) = (x + 3a)^2 + y^2 \]

6. **Expanding and Simplifying**:
- Expanding both sides:
\[ 4(x^2 - 6ax + 9a^2 + y^2) = x^2 + 6ax + 9a^2 + y^2 \]
- This simplifies to:
\[ 4x^2 - 24ax + 36a^2 + 4y^2 = x^2 + 6ax + 9a^2 + y^2 \]
- Rearranging gives:
\[ 3x^2 - 30ax + 27a^2 + 3y^2 = 0 \]
- Dividing through by 3:
\[ x^2 - 10ax + 9a^2 + y^2 = 0 \]

7. **Completing the Square**:
- Completing the square for the \(x\) terms:
\[ (x - 5a)^2 + y^2 = 16a^2 \]
- This represents a circle centered at \((5a, 0)\) with a radius of \(4a\).

8. **Conclusion**:
- The points where the electric potential is zero lie on a circle with center at \((5a, 0)\) and radius \(4a\).

### Correct Statements:
- The first option is correct: The point where the electric potential due to the two charges is zero in the xy-plane lies on a circle of radius \(4a\) centered at \((5a, 0)\).
- The second option is also correct: The potential is zero at \(x = a\) and \(x = 9a\) since both points lie on the circle.