Let a circle pass through the origin and its centre be the point of...

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(\dfrac{-1}{k-1}\right) \left(\dfrac{-2}{k^2}\right) = -1$<br><br>$\dfrac{2}{k^2(k-1)} = -1 \Rightarrow k^3 - k^2 + 2 = 0$<br><br>By trial,$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$.<br><br>Substituting$k = -1$into the equations of the lines, we get:<br><br>$x - 2y + 3 = 0$<br><br>$2x + y - 4 = 0$<br><br>Solving these two equations, we get the point of intersection as$x = 1 and y = 2$. Thus, the centre of the circle is$C(1, 2)$.<br><br>Since the circle passes through the origin$(0,0)$, the square of its radius$R$is:<br><br>$R^2 = (1 - 0)^2 + (2 - 0)^2 = 5$<br><br>The perpendicular distance$d$from the centre$C(1, 2)$to the line$x - y + 2 = 0$is:<br><br>$d = \dfrac{|1 - 2 + 2|}{\sqrt{1^2 + (-1)^2}} = \dfrac{1}{\sqrt{2}}$<br><br>The length of the chord$AB$is given by$2\sqrt{R^2 - d^2}$. Therefore, its square is:<br><br>$(AB)^2 = 4(R^2 - d^2) = 4\left(5 - \dfrac{1}{2}\right) = 4 \times \dfrac{9}{2} = 18$<br><br>Answer:$18