Square of Length of Chord Intersected by Circle

To find the value of $(AB)^2$, we proceed step-by-step:

Step 1: Determine the value of $k$ The given lines are: $L_1: x + (k - 1)y + 3 = 0$ $L_2: 2x + k^2y - 4 = 0$

Since $L_1$ and $L_2$ are mutually perpendicular, the product of their coefficients of $x$ plus the product of their coefficients of $y$ must be equal to $0$ ($a_1a_2 + b_1b_2 = 0$): $(1)(2) + (k - 1)(k^2) = 0$ $2 + k^3 - k^2 = 0$ $k^3 - k^2 + 2 = 0$

Factoring this cubic equation, we get: $(k + 1)(k^2 - 2k + 2) = 0$

Since $k^2 - 2k + 2 = 0$ has no real roots (as its discriminant $D = (-2)^2 - 4(1)(2) = -4 < 0$), the only real root is: $k = -1$

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Step 2: Find the centre of the circle Substituting $k = -1$ into the equations of $L_1$ and $L_2$: $L_1: x - 2y + 3 = 0$ $L_2: 2x + y - 4 = 0$

To find their point of intersection (the centre $C(h, k)$ of the circle), we solve these equations simultaneously: From $L_2$, we have $y = 4 - 2x$. Substituting this into $L_1$: $x - 2(4 - 2x) + 3 = 0$ $x - 8 + 4x + 3 = 0$ $5x - 5 = 0 \implies x = 1$

Thus, $y = 4 - 2(1) = 2$. So, the centre of the circle is $C(1, 2)$.

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Step 3: Find the radius of the circle The circle passes through the origin $O(0, 0)$. Therefore, the radius $R$ is the distance between the centre $C(1, 2)$ and $O(0, 0)$: $R^2 = (1 - 0)^2 + (2 - 0)^2 = 1 + 4 = 5$

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Step 4: Calculate the length of the chord $(AB)^2$ The equation of the line intersecting the circle is $x - y + 2 = 0$.

The perpendicular distance $d$ from the centre $C(1, 2)$ to this line is: $d = \frac{|1 - 2 + 2|}{\sqrt{1^2 + (-1)^2}} = \frac{1}{\sqrt{2}}$

The length of the chord $AB$ is given by: $AB = 2\sqrt{R^2 - d^2}$

Squaring both sides: $(AB)^2 = 4\left(R^2 - d^2\right)$

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

Correct Option: C