If $$a$$ and $$c$$ are positive real numbers and the ellipse $$\frac{x^2}{4c^2} + \frac{y^2}{c^2} = 1$$ has four distinct points in common with the circle $$x^2 + y^2 = 9a^2$$, then

The ellipse $$\frac{x^2}{4c^2} + \frac{y^2}{c^2} = 1$$ has semi-major axis $$2c$$ (along the $$x$$-axis) and semi-minor axis $$c$$ (along the $$y$$-axis). The circle $$x^2 + y^2 = 9a^2$$ has radius $$3a$$.

For the circle and ellipse to have four distinct intersection points, the circle's radius must lie strictly between the semi-minor and semi-major axes of the ellipse. That is, $$c < 3a < 2c$$.

From $$3a > c$$, we get $$9a^2 > c^2$$, i.e., $$c^2 < 9a^2$$. From $$3a < 2c$$, we get $$c > 3a/2$$, i.e., $$c^2 > 9a^2/4$$, which gives $$4c^2 > 9a^2$$, or equivalently $$9a^2 - 4c^2 < 0$$.

Now we check which option is implied by these two conditions: $$3a/2 < c < 3a$$. Consider the expression $$9ac - 9a^2 - 2c^2$$. We can write this as $$-(2c^2 - 9ac + 9a^2) = -(2c - 3a)(c - 3a)$$. Under our conditions, $$c > 3a/2$$ means $$2c > 3a$$, so $$2c - 3a > 0$$. Also $$c < 3a$$ means $$c - 3a < 0$$. Therefore $$(2c - 3a)(c - 3a) < 0$$, and so $$-(2c - 3a)(c - 3a) > 0$$. This gives $$9ac - 9a^2 - 2c^2 > 0$$.