The radius of a circle centered at (0, 3) and | vedclass

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(A) For the ellipse $\frac{x^2}{16} + \frac{y^2}{9} = 1$,we have $a^2 = 16$ and $b^2 = 9$.The eccentricity $e$ is given by $e = \sqrt{1 - \frac{b^2}{a

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$4$

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(A) For the ellipse $\frac{x^2}{16} + \frac{y^2}{9} = 1$,we have $a^2 = 16$ and $b^2 = 9$.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 are at $(\pm ae, 0)$,which are $(\pm 4 	imes \frac{\sqrt{7}}{4}, 0) = (\pm \sqrt{7}, 0)$.The circle is centered at $(0, 3)$ and passes through the foci $(\sqrt{7}, 0)$ and $(-\sqrt{7}, 0)$.The radius $r$ is the distance between $(0, 3)$ and $(\sqrt{7}, 0)$.$r = \sqrt{(\sqrt{7} - 0)^2 + (0 - 3)^2} = \sqrt{7 + 9} = \sqrt{16} = 4$.

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

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