JEE Main 2026 — Apr 2, Shift 2 — Mathematics solution

JEE Main2026Apr 2, Shift 2Mathematics

Let a circle pass through the origin and its centre be the point of intersection of two mutually perpendicular lines $x + (k - 1) y + 3 = 0$ and $2 x + k^{2} y - 4 = 0$ . If the line $x - y + 2 = 0$ intersects the circle at the points $A$ and $B$ , then $(A B)^{2}$ is equal to :

A
$10$
B
$27$
C
$18$
D
$34$

Solution

The slopes of the two lines are $\frac{- 1}{k - 1}$ and $\frac{- 2}{k ^{2}}$ . Perpendicularity gives $\frac{2}{k ^{2} ( k - 1 )} = - 1$ , leading to $k^{3} - k^{2} + 2 = 0$ , which factorises so that $k = - 1$ .

With $k = - 1$ , the lines become $x - 2 y + 3 = 0$ and $2 x + y - 4 = 0$ . Solving these gives the centre $(1, 2)$ .

Since the circle passes through the origin, $r = 1^{2} + 2^{2} = 5$ .

Perpendicular distance from $(1, 2)$ to the line $x - y + 2 = 0$ is $p = \frac{∣1 - 2 + 2∣}{2} = \frac{1}{2}$ .

By the chord length formula, $A B = 2 r^{2} - p^{2} = 2 5 - \frac{1}{2} = 2 \cdot \frac{3}{2} = \frac{6}{2}$ .

$∴ (A B)^{2} = \frac{36}{2} = 18$ .

← Back to full paper

Practice more Mathematics

Concept-wise practice with instant solutions on Fundamenthol.

Start practicing →