JEE Advanced 2026 — Paper 2 — PHY-II solution

JEE Advanced2026Paper 2PHY-II

Two charges $Q_{1} = q$ and $Q_{2} = m q$ are placed at the points $P_{1} (a, b)$ and $P_{2} (ma, mb)$ , respectively, in the $X Y$ plane, where $a, b \neq = 0$ and $m \neq = 0, 1$ . If $V_{1}$ is the potential at a point in the $X Y$ 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:

A
For $m = - 1$ , locus of these points is $a x + b y = 0$ .
B
For $m = 2$ , the locus of these points is a circle of radius $\frac{2}{3} a^{2} + b^{2}$ centered at $(\frac{2}{3} a, \frac{2}{3} b)$
C
For $m = - 2$ , the locus of these points is a circle of radius $2 a^{2} + b^{2}$ centered at $(2 a, 2 b)$
D
For $m = - 3$ , locus of these points is $3 b x + 3 a y = 0$ .

Solution

Setup. Let $P (x, y)$ satisfy $∣ V_{1} ∣ = ∣ V_{2} ∣$ . With $V_{1} = \frac{k q}{r _{1}}$ and $V_{2} = \frac{k m q}{r _{2}}$ where $r_{1}^{2} = (x - a)^{2} + (y - b)^{2}$ and $r_{2}^{2} = (x - ma)^{2} + (y - mb)^{2}$ , the condition becomes $r_{2} = ∣ m ∣ r_{1}$ , i.e. $r_{2}^{2} = m^{2} r_{1}^{2}$ .

Expand and simplify. Subtracting:

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

$\Rightarrow (1 - m^{2}) (x^{2} + y^{2}) + 2 (m^{2} - m) (a x + b y) = 0$

$\Rightarrow (m + 1) (x^{2} + y^{2}) = 2 m (a x + b y)$ .

(A) $m = - 1$ . LHS $= 0$ , so $a x + b y = 0$ . Correct.

(B) $m = 2$ . $3 (x^{2} + y^{2}) = 4 (a x + b y) \Rightarrow x^{2} + y^{2} - \frac{4 a}{3} x - \frac{4 b}{3} y = 0$ . Centre $(\frac{2 a}{3}, \frac{2 b}{3})$ , radius $\frac{2}{3} a^{2} + b^{2}$ . Correct.

(C) $m = - 2$ . $- (x^{2} + y^{2}) = - 4 (a x + b y) \Rightarrow x^{2} + y^{2} = 4 (a x + b y)$ . Centre $(2 a, 2 b)$ , radius $2 a^{2} + b^{2}$ . Correct.

(D) $m = - 3$ . $- 2 (x^{2} + y^{2}) = - 6 (a x + b y) \Rightarrow x^{2} + y^{2} = 3 (a x + b y)$ , a circle, not the line $3 b x + 3 a y = 0$ . Incorrect.

$∴$ Correct options are (A), (B), (C).

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