Let the equation of an ellipse be frac{x^{2}} | vedclass

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(C) The given equation of the ellipse is $\frac{x^{2}}{144}+\frac{y^{2}}{25}=1$.Here, $a^{2}=144$ and $b^{2}=25$.The eccentricity $e$ is given by $e=\

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$11$

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(C) The given equation of the ellipse is $\frac{x^{2}}{144}+\frac{y^{2}}{25}=1$.Here, $a^{2}=144$ and $b^{2}=25$.The eccentricity $e$ is given by $e=\sqrt{1-\frac{b^{2}}{a^{2}}}=\sqrt{1-\frac{25}{144}}=\sqrt{\frac{119}{144}}=\frac{\sqrt{119}}{12}$.The foci of the ellipse are $(\pm ae, 0) = (\pm 12 	imes \frac{\sqrt{119}}{12}, 0) = (\pm \sqrt{119}, 0)$.The circle has its center at $(0, \sqrt{2})$ and passes through the foci $(\pm \sqrt{119}, 0)$.The radius $r$ of the circle is the distance between the center $(0, \sqrt{2})$ and one of the foci $(\sqrt{119}, 0)$.$r = \sqrt{(\sqrt{119}-0)^{2} + (0-\sqrt{2})^{2}}$$r = \sqrt{119 + 2} = \sqrt{121} = 11$.

Let the equation of an ellipse be $\frac{x^{2}}{144}+\frac{y^{2}}{25}=1$. Then, the radius of the circle with center $(0, \sqrt{2})$ and passing through the foci of the ellipse is

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