Question

Sol. [Key Concept: Points where two lines meet the axes are concyclic ⇒ (product of the $x$-intercepts) $=$ (product of the $y$-intercepts) — both equal the circle's constant term $c$.]

Possible cases are shown below

intercepts —

$L_{1:3}x - 2y - 12 = 0$ gives $A(4,0),\ B(0, - 6)$;

$L_{2:}x + ky + 3 = 0$ gives $C( - 3,0),\ D(0, - 3/k)$.

Circle through $A,C$ (on the $x$-axis) and $B$: $x^{2 +}y^{2 -}x + 4y - 12 = 0$.

Its $y$-axis points $(x = 0)$: $y^{2 + 4}y - 12 = 0 \Rightarrow \ y = 2,\ - 6$.

The fourth point $D(0, - 3/k)$ must be one of them:

$- 3/k = 2 \Rightarrow \ k = - \frac{3}{2}$, or $- 3/k = - 6 \Rightarrow \ k = \frac{1}{2}$ (here $D$ coincides with $B$).

Hence 2 values of k → (b).