Question:

Two charges $Q_1 = q$ and $Q_2 = mq$ are placed at the points $P_1(a, b)$ and $P_2(ma, mb)$, respectively, in the $XY$ plane, where $a, b \neq 0$ and $m \neq 0, 1$. If $V_1$ is the potential at a point in the $XY$ plane due to charge $Q_1$ and $V_2$ is the potential at that point due to charge $Q_2$. Correct statement(s) for the points at which $|V_1| = |V_2|$ is/are:

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The locus of $|V_1| = |V_2|$ for any two charges is always a circle known as the Circle of Apollonius, except when $|Q_1| = |Q_2|$, in which case it is a straight line.
Updated On: May 20, 2026
  • For $m = -1$, locus of these points is $ax + by = 0$.
  • For $m = 2$, the locus of these points is a circle of radius $\frac{2}{3}\sqrt{a^2 + b^2}$ centered at $(\frac{2}{3}a, \frac{2}{3}b)$.
  • For $m = -2$, the locus of these points is a circle of radius $2\sqrt{a^2 + b^2}$ centered at $(2a, 2b)$.
  • For $m = -3$, locus of these points is $3ax + 3by = 0$.
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The Correct Option is A

Solution and Explanation

Step 1: Understanding the Question:
The condition $|V_1| = |V_2|$ implies $\frac{k|Q_1|}{r_1} = \frac{k|Q_2|}{r_2}$. This relates the distances from the two fixed charges $P_1$ and $P_2$.

Step 2: Key Formula or Approach:


• Distance relation: $\frac{|q|}{r_1} = \frac{|mq|}{r_2} \implies r_2 = |m| r_1$.

• Apollonius Circle: The locus of points whose distances from two fixed points are in a constant ratio is a circle (or a line if the ratio is 1).

Step 3: Detailed Explanation:


Case $m = -1$: $r_2 = |-1| r_1 = r_1$. This is the perpendicular bisector of $P_1(a, b)$ and $P_2(-a, -b)$.
$(x-a)^2 + (y-b)^2 = (x+a)^2 + (y+b)^2 \implies 4ax + 4by = 0 \implies ax + by = 0$. (A) is correct.

Case $m = 2$:
$r_2 = 2r_1 \implies (x-2a)^2 + (y-2b)^2 = 4[(x-a)^2 + (y-b)^2]$.
$x^2 - 4ax + 4a^2 + y^2 - 4by + 4b^2 = 4x^2 - 8ax + 4a^2 + 4y^2 - 8by + 4b^2$.
$3x^2 - 4ax + 3y^2 - 4by = 0 \implies x^2 - \frac{4}{3}ax + y^2 - \frac{4}{3}by = 0$.
Center is $(2a/3, 2b/3)$. Radius squared $R^2 = (2a/3)^2 + (2b/3)^2 = \frac{4}{9}(a^2 + b^2) \implies R = \frac{2}{3}\sqrt{a^2 + b^2}$. (B) is correct.

Step 4: Final Answer:

The correct statements are (A) and (B).
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