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(B) Given the equation of the ellipse: $\frac{x^2}{16}+\frac{y^2}{9}=1$. Here,$a^2=16$ and $b^2=9$,so $a=4$ and $b=3$. Since $a > b$,the eccentricity

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$4$

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(B) Given the equation of the ellipse: $\frac{x^2}{16}+\frac{y^2}{9}=1$. Here,$a^2=16$ and $b^2=9$,so $a=4$ and $b=3$. Since $a > b$,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 $(\pm ae, 0) = (\pm 4 	imes \frac{\sqrt{7}}{4}, 0) = (\pm \sqrt{7}, 0)$. The circle passes through $(\sqrt{7}, 0)$ and $(-\sqrt{7}, 0)$ with its centre at $(0, 3)$. The radius $r$ is the distance between the centre $(0, 3)$ 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 centre at $(0,3)$ is

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