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 $=(\pm ae,0)=(\pm \sqrt{7},0)$
Since, the circle passes through $(\pm \sqrt{7},0)$ and centre at $(0,3)$.
$\therefore$ Radius $=\sqrt{(0-\sqrt{7}{)}^2+(3-0{)}^2}$
$=\sqrt{7+9}=\sqrt{16}=4$