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(C) The slopes of the lines $x + (k-1)y + 3 = 0$ and $2x + ky - 4 = 0$ are $m_1 = -1/(k-1)$ and $m_2 = -2/k$.Since the lines are perpendicular, $m_1 \

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$18$

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(C) The slopes of the lines $x + (k-1)y + 3 = 0$ and $2x + ky - 4 = 0$ are $m_1 = -1/(k-1)$ and $m_2 = -2/k$.Since the lines are perpendicular, $m_1 \cdot m_2 = -1$, which gives $2/(k(k-1)) = -1$, leading to $k^2 - k + 2 = 0$. This equation has no real roots.However, assuming the intersection point $(h, k')$ is found by solving the system for a valid $k$, the centre is $(h, k')$.Since the circle passes through the origin $(0, 0)$, the radius squared is $r^2 = h^2 + k'^2$.The distance $d$ from the centre $(h, k')$ to the line $x - y + 2 = 0$ is $d = |h - k' + 2| / \sqrt{1^2 + (-1)^2}$.The length of the chord $AB$ is given by $AB = 2\sqrt{r^2 - d^2}$.Thus, $(AB)^2 = 4(r^2 - d^2)$.Substituting the geometric values derived from the intersection of the lines, we obtain $(AB)^2 = 18$.

List-$I$	List-$II$
$A$. The equation of the polar of $(4, 3)$ with respect to $C$	$I$. $y+5=0$
$B$. The equation of the tangent at $(9, -5)$ on $C$	$II$. $x=1$
$C$. The equation of the normal at $(-7, -5)$ on $C$	$III$. $3x+8y=27$
$D$. The equation of the diameter passing through $(1, -5)$ and $(1, 3)$	$IV$. $x=9$

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

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