JEE Main 2026 — Mathematics Coordinate Geometry

The given lines are $x + (k-1)y + 3 = 0$x+(k−1)y+3=0 and $2x + k^2 y - 4 = 0$2x+k2y−4=0.

Since they are mutually perpendicular, the product of their slopes is $-1$−1:

$$\left(\dfrac{-1}{k-1}\right) \left(\dfrac{-2}{k^2}\right) = -1$$(k−1−1​)(k2−2​)=−1

$$\dfrac{2}{k^2(k-1)} = -1 \Rightarrow k^3 - k^2 + 2 = 0$$k2(k−1)2​=−1⇒k3−k2+2=0

By trial, $k = -1$k=−1 is a root. Factoring gives $(k+1)(k^2 - 2k + 2) = 0$(k+1)(k2−2k+2)=0. Since $k^2 - 2k + 2 = 0$k2−2k+2=0 has no real roots, $k = -1$k=−1.

Substituting $k = -1$k=−1 into the equations of the lines, we get:

$$x - 2y + 3 = 0$$x−2y+3=0

$$2x + y - 4 = 0$$2x+y−4=0

Solving these two equations, we get the point of intersection as $x = 1$x=1 and $y = 2$y=2. Thus, the centre of the circle is $C(1, 2)$C(1,2).

Since the circle passes through the origin $(0,0)$(0,0), the square of its radius $R$R is:

$$R^2 = (1 - 0)^2 + (2 - 0)^2 = 5$$R2=(1−0)2+(2−0)2=5

The perpendicular distance $d$d from the centre $C(1, 2)$C(1,2) to the line $x - y + 2 = 0$x−y+2=0 is:

$$d = \dfrac{|1 - 2 + 2|}{\sqrt{1^2 + (-1)^2}} = \dfrac{1}{\sqrt{2}}$$d=12+(−1)2​∣1−2+2∣​=2​1​

The length of the chord $AB$AB is given by $2\sqrt{R^2 - d^2}$2R2−d2​. Therefore, its square is:

$$(AB)^2 = 4(R^2 - d^2) = 4\left(5 - \dfrac{1}{2}\right) = 4 \times \dfrac{9}{2} = 18$$(AB)2=4(R2−d2)=4(5−21​)=4×29​=18

Answer: $18$18