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Q.
The radius of the circle passing through the foci of the ellipse $9 x^{2}+16 y^{2}=144$ and having its centre at $(0,3)$, is
ManipalManipal 2015
Solution:
We have,
$\frac{x^{2}}{16}+\frac{y^{2}}{9}=1$
The eccentricity $e$ of the ellipse is given by
$\theta=\sqrt{1-\frac{9}{16}}=\frac{\sqrt{7}}{4}$
So, the coordinates of the foci are $(\pm \sqrt{7}, 0)$.
$\therefore $ Radius of the circle
$=\sqrt{(\sqrt{7}-0)^{2}+(0-3)^{2}} $
$=\sqrt{7+9}=\sqrt{16}=4$