Two charges Q₁ = q and Q₂ = mq are placed at the points P₁(a, b) and P₂(ma, mb), respectively, in the X Y plane, where a, b ≠ 0 and m ≠ 0, 1. If V₁ is the potential at a point in the X Y plane due to charge Q₁ and V₂ is the potential at that point due to charge Q₂. Correct statement(s) for the points at which |V₁| = |V₂| is/are:

Step 1: Express the condition for electric potential magnitude.
Let any arbitrary point in the XY-plane be (x, y).
The distance of (x, y) from point P₁(a, b) carrying charge Q₁ = q is given by:

$r_1=\sqrt{{(x-a)}^2+{(y-b)}^2}$

Similarly, the distance of (x, y) from point P₂(ma, mb) carrying charge Q₂ = mq is given by:

$r_2=\sqrt{{(x-ma)}^2+{(y-mb)}^2}$

The electric potential due to a point charge Q at a distance r is $V=\frac{1}{4\pi {\epsilon }_0}\frac{Q}{r}$.
We are given that $\left|V_1\right|=\left|V_2\right|$, which leads to:

$\frac{\left|q\right|}{r_1}=\frac{\left|mq\right|}{r_2}\Rightarrow r_2=\left|m\right|r_1$

Squaring both sides gives the general equation for the locus:

${(x-ma)}^2+{(y-mb)}^2=m^2[{(x-a)}^2+{(y-b)}^2]$

Step 2: Case 1: Analyze for m = -1
Substitute m = -1 into the distance condition $r_2=|-1|r_1=r_1$:

${(x+a)}^2+{(y+b)}^2={(x-a)}^2+{(y-b)}^2$

Expanding both sides:

$x^2+2ax+a^2+y^2+2by+b^2=x^2-2ax+a^2+y^2-2by+b^2$

Simplifying by canceling common terms $x^2+y^2+a^2+b^2$:

$4ax+4by=0\Rightarrow ax+by=0$

Hence, for m = -1, the locus is a straight line given by ax + by = 0.

Step 3: Case 2: Analyze for m = 2
Substitute m = 2 into the squared distance equation:

${(x-2a)}^2+{(y-2b)}^2=4[{(x-a)}^2+{(y-b)}^2]$

Expanding both sides:

$x^2-4ax+4a^2+y^2-4by+4b^2=4x^2-8ax+4a^2+4y^2-8by+4b^2$

Rearranging all terms to one side:

$3x^2-4ax+3y^2-4by=0$

Dividing by 3:

$x^2-\frac{4}{3}ax+y^2-\frac{4}{3}by=0$

Completing the squares for x and y:

${(x-\frac{2}{3}a)}^2+{(y-\frac{2}{3}b)}^2=\frac{4}{9}{a}^2+\frac{4}{9}{b}^2={\left(\frac{2}{3}\sqrt{a^2+b^2}\right)}^2$

This represents a circle centered at $(\frac{2}{3}a,\frac{2}{3}b)$ with radius $R=\frac{2}{3}\sqrt{a^2+b^2}$.

Thus, both the statement for m = -1 and the statement for m = 2 are fully verified.