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(C) The equation of the ellipse is $\frac{x^2}{(2c)^2} + \frac{y^2}{c^2} = 1$. The vertices are at $(\pm 2c, 0)$ and $(0, \pm c)$.The equation of the

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$9ac - 9a^2 - 2c^2 > 0$

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(C) The equation of the ellipse is $\frac{x^2}{(2c)^2} + \frac{y^2}{c^2} = 1$. The vertices are at $(\pm 2c, 0)$ and $(0, \pm c)$.The equation of the circle is $x^2 + y^2 = (3a)^2$. The radius is $3a$.For the circle and ellipse to have four distinct points of intersection,the radius of the circle must be greater than the semi-minor axis of the ellipse and less than the semi-major axis of the ellipse.Thus,$c From $3a From $c  0$.Substituting $y^2 = 9a^2 - x^2$ into the ellipse equation: $\frac{x^2}{4c^2} + \frac{9a^2 - x^2}{c^2} = 1$.$x^2 + 4(9a^2 - x^2) = 4c^2$ $\Rightarrow x^2 + 36a^2 - 4x^2 = 4c^2$ $\Rightarrow 3x^2 = 36a^2 - 4c^2$ $\Rightarrow x^2 = 12a^2 - \frac{4}{3}c^2$.For two distinct values of $x^2$ (which give four points),we need $0 $0 $12a^2 - \frac{4}{3}c^2 > 0$ $\Rightarrow 36a^2 > 4c^2$ $\Rightarrow 9a^2 > c^2$ $\Rightarrow 3a > c$.$12a^2 - \frac{4}{3}c^2 Combining these,$c  0$ is derived from the intersection analysis.

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

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