JEE Main 2026 · 2 Apr · Shift 2 · Mathematics Q10

← JEE Main 2026 · 2 Apr · Shift 2

MathematicsSingle correct+4 / -1Question 10 of 75

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 $2x+k^2y-4=0$ . If the line $x-y+2=0$ intersects the circle at the points A and B, then $(AB)^2$ is equal to :

Answer and worked solution

Official answer: C

Perpendicular lines. $1\cdot2+(k-1)k^2=0$ , i.e. $k^3-k^2+2=0$ , so $k=-1$ . The lines $x-2y+3=0$ and $2x+y-4=0$ meet at $(1,2)$ .

Circle. Centre $(1,2)$ through the origin: $r^2=5$ .

Chord. The distance from the centre to $x-y+2=0$ is $\frac1{\sqrt2}$ , so
$AB^2=4\left(5-\tfrac12\right)=18.$

(C)

Topics:Equation of a circle Angle between lines

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JEE Main 2026 · 2 Apr · Shift 2, Mathematics question 10