JEE Main · 2026 · 04 Apr 2026 · Shift 1 · Circle

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 RS is

(α, β)

, then

6 (α + β)

is equal to :

(A)1
(B)3✓ Correct
(C)4
(D)6

Answer: (B)

Solution

The point

P (1, 2)

lies on the circle since

1 + 4 + 1 - 6 = 0

. A chord from

P

whose midpoint is

(0, λ)

on the

y

-axis has its other end

S = 2 (0, λ) - (1, 2) = (- 1, 2 λ - 2)

. Requiring

S

x^{2} + y^{2} + x - 3 y = 0

gives

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

, i.e.

(2 λ - 2) (2 λ - 5) = 0

, so

λ = 1

λ = \frac{5}{2}

. The two other ends are

(- 1, 0)

and

(- 1, 3)

, whose midpoint is

(α, β) = (- 1, \frac{3}{2})

. Thus

6 (α + β) = 6 (- 1 + \frac{3}{2}) = 3

JEE Main · 2026 · 04 Apr 2026 · Shift 1 · Single correct option

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