JEE Main 2026 Apr 4 Shift 1, Mathematics Q13: Circle: Chord, Tangent And Normal

JEE Main2026Apr 4, Shift 1Mathematics

Suppose the two chords, drawn from the point $(1, 2)$ on the circle $x^{2} + y^{2} + x - 3 y = 0$ are bisected by the y-axis. If the other ends of these chords are $R$ and $S$ , and the mid point of the line segment $R S$ is $(α, β)$ , then $6 (α + β)$ is equal to :

A
$1$
B
$3$
C
$4$
D
$6$

Solution

Let $P = (1, 2)$ and let the chord from $P$ be bisected by the $y$ -axis at $M = (0, λ)$ . The other endpoint $T = 2 M - P = (- 1, 2 λ - 2)$ must lie on the circle:

$(- 1)^{2} + (2 λ - 2)^{2} + (- 1) - 3 (2 λ - 2) = 0.$

$\Rightarrow 4 λ^{2} - 8 λ + 4 - 6 λ + 6 = 0 \Rightarrow 4 λ^{2} - 14 λ + 10 = 0 \Rightarrow 2 λ^{2} - 7 λ + 5 = 0.$

Roots: $λ = 1$ or $λ = \frac{5}{2}$ , giving the two chord endpoints

$S = (- 1, 0) and R = (- 1, 3) .$

Midpoint of $R S$ : $(- 1, \frac{3}{2})$ , so $α = - 1, β = \frac{3}{2}$ and

$6 (α + β) = 6 \cdot \frac{1}{2} = 3.$

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JEE Main 2026 Apr 4 Shift 1, Mathematics Q13: Circle: Chord, Tangent And Normal

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