JEE Main 2026 — 2 April, Evening Shift — Question 32

$x+(k-1) y+3=0$ $2 \mathrm{x}+\mathrm{k}^{2} \mathrm{y}-4=0$ $\left(\frac{1}{1-\mathrm{k}}\right)\left(\frac{2}{\mathrm{k}^{2}}\right)=1$ $2=\mathrm{k}^{2}-\mathrm{k}^{3}$ $\mathrm{k}^{3}-\mathrm{k}^{2}+2=0$ $\mathrm{k}=-1$ Solving : $2(\mathrm{x}-2 \mathrm{y}+3)=0$ $2 \mathrm{x}+\mathrm{y}-4=0$ $-5 \mathrm{y}+10=0$ $\mathrm{y}=2$ $\mathrm{x}=1$ Centre $(1,2)$ $\mathrm{r}=\sqrt{5}$ So circle is $(x-1)^{2}+(y-2)^{2}=5$ Chord $\mathrm{x}-\mathrm{y}+2=0$ $\mathrm{p}=\frac{1}{\sqrt{2}}$ $\ell=\sqrt{5-\frac{1}{2}}=\frac{3}{\sqrt{2}}$ $\mathrm{AB}=\frac{6}{\sqrt{2}} \therefore \mathrm{AB}^{2}=18$