Question

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

• A. 9
• B. 7
• C. 11
• D. 5

Answer 1

Answer: (C)

Given equation of ellipse is
$\frac{x^{2}}{144}+\frac{y^{2}}{25} =1 $
Here, $a^{2} =144 $ and $ b^{2}=25$
Now,$e=\sqrt{1-\frac{b^{2}}{a^{2}}}=\sqrt{1-\frac{25}{144}}$
$=\sqrt{\frac{119}{144}}=\frac{\sqrt{119}}{12}$
$\therefore $ Foci of an ellipse $=(\pm a e, 0)$
$=\left(\pm 12 \times \frac{\sqrt{119}}{12}, 0\right)$
$=(\pm \sqrt{119}, 0)$
Since, the circle with centre $(0, \sqrt{2})$ and passing through foci $(\pm \sqrt{119}, 0)$ of the ellipse.
$\therefore $ Radius of circle $=\sqrt{(\sqrt{119}-0)^{2}+(0-\sqrt{2})^{2}}$
$=\sqrt{119+2}=\sqrt{121}=11$