Let a circle $$C$$ touch the lines $$L_1: 4x - 3y + K_1 = 0$$ and $$L_2: 4x - 3y + K_2 = 0$$, $$K_1, K_2 \in R$$. If a line passing through the centre of the circle $$C$$ intersects $$L_1$$ at $$(-1, 2)$$ and $$L_2$$ at $$(3, -6)$$, then the equation of the circle $$C$$ is

We need to find the equation of circle $$C$$ that touches the parallel lines $$L_1$$ and $$L_2$$. To determine the constants, note that point $$(-1, 2)$$ lies on $$L_1: 4x - 3y + K_1 = 0$$, so $$4(-1) - 3(2) + K_1 = 0 \implies K_1 = 10$$. Similarly, point $$(3, -6)$$ lies on $$L_2: 4x - 3y + K_2 = 0$$, so $$4(3) - 3(-6) + K_2 = 0 \implies K_2 = -30$$. Hence the equations are $$L_1: 4x - 3y + 10 = 0$$ and $$L_2: 4x - 3y - 30 = 0$$.

Since the circle touches both lines, its centre is equidistant from $$L_1$$ and $$L_2$$ and therefore lies on their midline, given by $$4x - 3y + \frac{10 + (-30)}{2} = 0 \implies 4x - 3y - 10 = 0$$.

The centre also lies on the line through $$(-1, 2)$$ and $$(3, -6)$$, which in parametric form is $$x = -1 + 4t, \quad y = 2 - 8t$$. Substituting into the midline equation $$4x - 3y - 10 = 0$$ gives $$4(-1 + 4t) - 3(2 - 8t) - 10 = 0$$, which simplifies to $$-4 + 16t - 6 + 24t - 10 = 0 \implies 40t - 20 = 0 \implies t = \tfrac{1}{2}$$. Thus the centre is $$\bigl(-1 + 4\cdot\tfrac{1}{2},\;2 - 8\cdot\tfrac{1}{2}\bigr) = (1, -2)\!.$$

The radius is the perpendicular distance from the centre $$(1, -2)$$ to $$L_1$$: $$r = \frac{\bigl|4(1) - 3(-2) + 10\bigr|}{\sqrt{16 + 9}} = \frac{|4 + 6 + 10|}{5} = \frac{20}{5} = 4$$. Verification with $$L_2$$ yields $$\frac{\bigl|4(1) - 3(-2) - 30\bigr|}{5} = 4$$ ✓.

Therefore, the equation of the circle is $$(x - 1)^2 + (y + 2)^2 = 4^2 = 16$$ and the correct answer is Option B: $$(x - 1)^2 + (y + 2)^2 = 16$$.