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(A) The equation of the ellipse is $\frac{x^2}{16} + \frac{y^2}{9} = 1$.Here,$a^2 = 16$ and $b^2 = 9$,so $a = 4$ and $b = 3$.The eccentricity $e$ is g

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$4$

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(A) The equation of the ellipse is $\frac{x^2}{16} + \frac{y^2}{9} = 1$.Here,$a^2 = 16$ and $b^2 = 9$,so $a = 4$ and $b = 3$.The eccentricity $e$ is given by $e = \sqrt{1 - \frac{b^2}{a^2}} = \sqrt{1 - \frac{9}{16}} = \sqrt{\frac{7}{16}} = \frac{\sqrt{7}}{4}$.The foci of the ellipse are $(\pm ae, 0) = (\pm 4 	imes \frac{\sqrt{7}}{4}, 0) = (\pm \sqrt{7}, 0)$.The circle has its center at $(0, 3)$ and passes through the foci $(\sqrt{7}, 0)$ and $(-\sqrt{7}, 0)$.The radius $r$ of the circle is the distance between the center $(0, 3)$ and one of the foci,say $(\sqrt{7}, 0)$.$r = \sqrt{(\sqrt{7} - 0)^2 + (0 - 3)^2} = \sqrt{7 + 9} = \sqrt{16} = 4$.

The radius of the circle passing through the foci of the ellipse $\frac{x^2}{16} + \frac{y^2}{9} = 1$ and having its center at $(0, 3)$ is:

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