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

[JEE Main 2026, 2 Apr (Shift 2)]

Question

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

[JEE Main 2026, 2 Apr (Shift 2)]

YESPYQ · Accepted Answer

Correct answer: c) $18$. $ x+(k-1) y+3=0 $ $ 2 x+k^2 y-4=0$ Since these $2$ lines are perpendicular $ \frac{1}{1-k}\left(-\frac{2}{k^2}\right)=-1 $ $ 2=k^2(1-k) $ $ \Rightarrow k^3-k^2+2=0$ $k^3-k^2+2=(k+1)\left(k^2-2 k+2\right)=0$, and since $k$ is real, $k=-1$ $∴$ Lines are : $x-2 y+3=0 $ and $2 x+y-4=0$ then $ x=1$ and $y=2$ $∴$ Centre is $(1,2)$ Circle will be $x^2+y^2-2 x-4 y=0$ Line $x-y+2=0$ will intersect at $A(-1,1)$ and $B(2,4)$ $ \therefore(A B)^2=(2+1)^2+(4-1)^2 $ $ =9+9=18$

Let a circle pass through the origin and its centre be the point of intersection of two mutually perpendicular lines \(x…

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