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)

$4$

We have, $\frac{x^2}{16}+\frac{y^2}{9}=1$

We know that, $b^2=a^2(1-e^2)$

$\Rightarrow$ $g=16$ $(1-e^2)$

$\Rightarrow$ $\frac{9}{16}=$ $(1-e^2)$

$\Rightarrow$ $e^2=\frac{16-9}{16}=\frac{7}{16}$

$\Rightarrow$ $e=\frac{\sqrt{7}}{4}$

$\therefore$ Focii $\to$ $\left(\sqrt{7,}0\right)$ & $(-\sqrt{7,}0)$

Circle $\Rightarrow$ $(x-0{)}^2+(y-3{)}^2=r^2$

$\Rightarrow$ $7+9=r^2$

$\Rightarrow$ $r=4$