Question

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

• A. $6$
• B. 4
• C. 3
• D. 2

Answer 1

Answer: (B)

We have, $\frac{x^{2}}{16}+\frac{y^{2}}{9}=1 \Rightarrow \frac{x^{2}}{4^{2}}+\frac{y^{2}}{3^{2}}=1$
$\therefore a=4, b=3$
Here, $a > b$
Now, $e=\sqrt{1-\frac{b^{2}}{a^{2}}}=\sqrt{1-\frac{9}{16}}=\frac{\sqrt{7}}{4}$
$\therefore $ Focus $=(\pm a e, 0)=(\pm \sqrt{7}, 0)$
Since, the circle passes through $(\pm \sqrt{7}, 0)$ and centre at $(0,3)$.
$\therefore $ Radius $=\sqrt{(0-\sqrt{7})^{2}+(3-0)^{2}}$
$=\sqrt{7+9}=\sqrt{16}=4$