JEE Advanced · 2026 · 17 May 2026 · Paper 2 · Electrostatics

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

Answer: (A), (B), (C)

Solution

The condition

∣ V_{1} ∣ = ∣ V_{2} ∣

gives

\frac{1}{r _{1}} = \frac{∣ m ∣}{r _{2}}

, so

r_{2}^{2} = m^{2} r_{1}^{2}

. With

r_{1}^{2} = (x - a)^{2} + (y - b)^{2}

and

r_{2}^{2} = (x - ma)^{2} + (y - mb)^{2}

, expanding and simplifying yields

(m + 1) (x^{2} + y^{2}) = 2 m (a x + b y)

.

(A) For

m = - 1

a x + b y = 0

. Correct.

(B) For

m = 2

3 (x^{2} + y^{2}) = 4 (a x + b y)

, which is a circle with center

(\frac{2 a}{3}, \frac{2 b}{3})

and radius

\frac{2}{3} a^{2} + b^{2}

. Correct.

(C) For

m = - 2

x^{2} + y^{2} = 4 (a x + b y)

, which is a circle with center

(2 a, 2 b)

and radius

2 a^{2} + b^{2}

. Correct.

(D) For

m = - 3

x^{2} + y^{2} = 3 (a x + b y)

, which is a circle, not a line. Incorrect.

JEE Advanced · 2026 · 17 May 2026 · Paper 2 · One or more correct options