For real numbers $$a$$, $$b$$ ($$a > b > 0$$), let
Area $$\{(x, y) : x^2 + y^2 \leq a^2$$ and $$\frac{x^2}{a^2} + \frac{y^2}{b^2} \geq 1\} = 30\pi$$
and
Area $$\{(x, y) : x^2 + y^2 \geq b^2$$ and $$\frac{x^2}{a^2} + \frac{y^2}{b^2} \leq 1\} = 18\pi$$
Then the value of $$(a-b)^2$$ is equal to ______.

We are given that for $$a > b > 0$$ the area of the set $$\{(x,y) : x^2 + y^2 \leq a^2 \text{ and } \frac{x^2}{a^2} + \frac{y^2}{b^2} \geq 1\}$$ is $$30\pi$$ and the area of the set $$\{(x,y) : x^2 + y^2 \geq b^2 \text{ and } \frac{x^2}{a^2} + \frac{y^2}{b^2} \leq 1\}$$ is $$18\pi$$.

The first region, which lies inside the circle of radius $$a$$ but outside the ellipse, has area $$\pi a^2 - \pi ab = \pi a(a - b) = 30\pi$$, giving $$a(a-b) = 30 \quad \cdots(1)$$. Similarly, the second region, which lies inside the ellipse but outside the circle of radius $$b$$, has area $$\pi ab - \pi b^2 = \pi b(a - b) = 18\pi$$, yielding $$b(a-b) = 18 \quad \cdots(2)$$.

On dividing equation (1) by equation (2), we obtain $$\dfrac{a}{b} = \dfrac{30}{18} = \dfrac{5}{3}$$, so $$a = \dfrac{5b}{3}$$. Substituting this into (2) leads to $$b\left(\dfrac{5b}{3} - b\right) = 18$$, hence $$b \cdot \dfrac{2b}{3} = 18 \Rightarrow b^2 = 27$$. It then follows that $$a - b = \dfrac{5b}{3} - b = \dfrac{2b}{3}$$.

Therefore, $$(a-b)^2 = \dfrac{4b^2}{9} = \dfrac{4 \times 27}{9} = \boxed{12}\,.$$