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

JEE Main · 2026 · 02 Apr 2026 · Shift 2 · Circle

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

(AB)^{2}

is equal to :

(A)10
(B)27
(C)18✓ Correct
(D)34

Answer: (C)

Solution

The lines are perpendicular when the product of their slopes is

- 1

\frac{- 1}{k - 1} \cdot \frac{- 2}{k ^{2}} = - 1

, which simplifies to

k^{3} - k^{2} + 2 = 0

with real root

k = - 1

. The lines become

x - 2 y + 3 = 0

and

2 x + y - 4 = 0

, meeting at the centre

(1, 2)

. As the circle passes through the origin,

r^{2} = 1^{2} + 2^{2} = 5

. The distance from

(1, 2)

x - y + 2 = 0

\frac{∣1 - 2 + 2∣}{2} = \frac{1}{2}

, so

A B = 2 r^{2} - d^{2} = 2 5 - \frac{1}{2} = 32,

giving

(A B)^{2} = 18

JEE Main · 2026 · 02 Apr 2026 · Shift 2 · Single correct option

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