JEE Main 2026 — Circle Question with Solution

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

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

$\left(\frac{-1}{k-1}\right)\left(\frac{-2}{k^2}\right)=-1$

$\frac{2}{k^2(k-1)}=-1\Rightarrow k^3-k^2+2=0$

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

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

$x-2y+3=0$

$2x+y-4=0$

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

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

$R^2=(1-0{)}^2+(2-0{)}^2=5$

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

$d=\frac{|1-2+2|}{\sqrt{1^2+(-1{)}^2}}=\frac{1}{\sqrt{2}}$

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

$(AB{)}^2=4(R^2-d^2)=4\left(5-\frac{1}{2}\right)=4\times \frac{9}{2}=18$

Answer: $18$