If the radius of the largest circle with centre (2, 0) inscribed in the ellipse $$x^2 + 4y^2 = 36$$ is $$r$$, then $$12r^2$$ is equal to _______

Given the ellipse $$x^2 + 4y^2 = 36$$, rewrite as $$\frac{x^2}{36} + \frac{y^2}{9} = 1$$ with $$a^2 = 36, b^2 = 9$$.

We need the largest circle centered at $$(2, 0)$$ inscribed in this ellipse. The circle equation is $$(x - 2)^2 + y^2 = r^2$$.

Substituting $$y^2 = r^2 - (x-2)^2$$ into the ellipse equation:

$$x^2 + 4[r^2 - (x-2)^2] = 36$$

$$x^2 + 4r^2 - 4x^2 + 16x - 16 = 36$$

$$-3x^2 + 16x + 4r^2 - 52 = 0$$

$$3x^2 - 16x + (52 - 4r^2) = 0$$

For the largest inscribed circle, the circle is tangent to the ellipse (touches at exactly one point), so the discriminant equals zero:

$$\Delta = 256 - 12(52 - 4r^2) = 0$$

$$256 - 624 + 48r^2 = 0$$

$$48r^2 = 368$$

$$r^2 = \frac{368}{48} = \frac{23}{3}$$

Therefore:

$$12r^2 = 12 \times \frac{23}{3} = 92$$

The answer is Option B: 92.