JEE Main 2 April 2026 Shift 2 Paper

Concept: The center of the circle is the intersection of two perpendicular lines. The square of the length of the chord is found using the distance from the center to the chord.Formula: Distance from point $(x_1, y_1)$ to line $Ax + By + C = 0$ is $\frac{\left| Ax_1 + By_1 + C \right|}{\sqrt{A^2 + B^2}}$. For a circle with center $(h, k)$ and radius $r$, and a chord of length $L$ at a distance $d$ from the center, $r^2 = d^2 + \left(\frac{L}{2}\right)^2$.Solution:The lines $x + (k - 1)y + 3 = 0$ and $2x + k^2y - 4 = 0$ are perpendicular.The condition for perpendicularity of lines $A_1x + B_1y + C_1 = 0$ and $A_2x + B_2y + C_2 = 0$ is $A_1A_2 + B_1B_2 = 0$.So, $1(2) + (k-1)k^2 = 0 \Rightarrow 2 + k^3 - k^2 = 0 \Rightarrow k^3 - k^2 + 2 = 0$.By inspection, $k = -1$ is a root: $(-1)^3 - (-1)^2 + 2 = -1 - 1 + 2 = 0$.The lines become $x - 2y + 3 = 0$ and $2x + y - 4 = 0$.Solving these for intersection: Multiply first by 2: $2x - 4y + 6 = 0$. Subtract from the second: $(2x + y - 4) - (2x - 4y + 6) = 0 \Rightarrow 5y - 10 = 0 \Rightarrow y = 2$.Substitute $y=2$ into $x - 2y + 3 = 0$: $x - 2(2) + 3 = 0 \Rightarrow x - 4 + 3 = 0 \Rightarrow x = 1$.The center of the circle is $(1, 2)$.Since the circle passes through the origin $(0, 0)$, the radius squared is $r^2 = (1-0)^2 + (2-0)^2 = 1^2 + 2^2 = 1 + 4 = 5$.The equation of the circle is $(x-1)^2 + (y-2)^2 = 5$, which is $x^2 - 2x + 1 + y^2 - 4y + 4 = 5 \Rightarrow x^2 + y^2 - 2x - 4y = 0$.The line $x - y + 2 = 0$ intersects the circle. The distance $d$ from the center $(1, 2)$ to this line is:$d = \frac{\left| 1 - 2 + 2 \right|}{\sqrt{1^2 + (-1)^2}} = \frac{\left| 1 \right|}{\sqrt{2}} = \frac{1}{\sqrt{2}}$.Let $AB$ be the length of the chord. Then $r^2 = d^2 + \left(\frac{AB}{2}\right)^2$.$5 = \left(\frac{1}{\sqrt{2}}\right)^2 + \left(\frac{AB}{2}\right)^2$$5 = \frac{1}{2} + \frac{(AB)^2}{4}$$5 - \frac{1}{2} = \frac{(AB)^2}{4}$$\frac{9}{2} = \frac{(AB)^2}{4}$$(AB)^2 = \frac{9}{2} \times 4 = 18$.Answer: 18