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 $=(\pm ae,0)$
$=\left(\pm 12\times \frac{\sqrt{119}}{12},0\right)$
$=(\pm \sqrt{119},0)$
Since, the circle with centre $(0,\sqrt{2})$ and passing through foci $(\pm \sqrt{119},0)$ of the ellipse.
$\therefore$ Radius of circle $=\sqrt{(\sqrt{119}-0{)}^2+(0-\sqrt{2}{)}^2}$
$=\sqrt{119+2}=\sqrt{121}=11$