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

The centre of the required circle is the intersection point of the two given lines. Because the lines are mutually perpendicular, the product of their slopes is $$-1$$.

First line: $$x + (k-1)y + 3 = 0 \;\Rightarrow\; y = -\dfrac{x}{k-1} - \dfrac{3}{k-1}$$ Hence slope $$m_1 = -\dfrac{1}{k-1}$$.

Second line: $$2x + k^{2}y - 4 = 0 \;\Rightarrow\; y = -\dfrac{2x}{k^{2}} + \dfrac{4}{k^{2}}$$ Hence slope $$m_2 = -\dfrac{2}{k^{2}}$$.

Perpendicular condition: $$m_1\,m_2 = -1$$ $$\left(-\dfrac{1}{k-1}\right)\!\left(-\dfrac{2}{k^{2}}\right) = -1 \;\Longrightarrow\; \dfrac{2}{k^{2}(k-1)} = -1 \;\Longrightarrow\; k^{2}(k-1) = -2 \;\Longrightarrow\; k^{3} - k^{2} + 2 = 0$$

Trying integral roots, $$k = -1$$ satisfies the equation: $$(-1)^{3} - (-1)^{2} + 2 = -1 -1 + 2 = 0.$$ (The remaining quadratic factor gives complex roots, so the real value is $$k = -1$$.)

Insert $$k = -1$$ into the two lines to find the centre $$C(x_c,y_c)$$:

Line 1: $$x - 2y + 3 = 0$$ Line 2: $$2x + y - 4 = 0$$

From line 2, $$y = 4 - 2x$$. Substituting into line 1: $$x - 2(4 - 2x) + 3 = 0 \;\Rightarrow\; 5x - 5 = 0 \;\Rightarrow\; x = 1.$$ Then $$y = 4 - 2(1) = 2.$$ Thus the centre is $$C(1,\,2).$$

The circle passes through the origin, so the radius squared is $$r^{2} = OC^{2} = 1^{2} + 2^{2} = 5.$$ Equation of the circle: $$(x-1)^{2} + (y-2)^{2} = 5.$$

The chord $$AB$$ is cut by the line $$x - y + 2 = 0 \;\Rightarrow\; y = x + 2.$$ Substitute into the circle’s equation:

$$(x-1)^{2} + (x + 2 - 2)^{2} = 5 \;\Longrightarrow\; (x-1)^{2} + x^{2} = 5 \;\Longrightarrow\; x^{2} - x - 2 = 0.$$

Solving, $$x = \dfrac{1 \pm \sqrt{1 + 8}}{2} = \dfrac{1 \pm 3}{2},$$ giving $$x_1 = 2,\; x_2 = -1.$$ Corresponding $$y$$ values (using $$y = x + 2$$): $$A(2,\,4), \; B(-1,\,1).$$

Distance squared between $$A$$ and $$B$$:

$$(AB)^{2} = (x_2 - x_1)^{2} + (y_2 - y_1)^{2} = (-1 - 2)^{2} + (1 - 4)^{2} = (-3)^{2} + (-3)^{2} = 9 + 9 = 18.$$

Hence $$(AB)^{2} = 18.$$

Option C which is: $$18$$