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 $(0,3)$ is

• A. $4$
• B. $\frac{3}{7}$
• C. $\sqrt{12}$
• D. $\frac{7}{2}$

Answer 1

The correct option is A

$4$

Calculating the radius of the circle:-

Given:-

Ellipse $\frac{x^2}{16}+\frac{y^2}{9}=1$ and its center$(0,3)$

Find:-

Radius of circle

The equation of the ellipse is $\frac{x^2}{16}+\frac{y^2}{9}=1$

$a=4$ and $b=3$

Eccentricity:-

$$\begin{gathered} e=\sqrt{1-\frac{b^2}{a^2}} \\ =\sqrt{1-\frac{9}{16}} \\ =\sqrt{\frac{7}{16}} \\ =\frac{\sqrt{7}}{4} \end{gathered}$$

Foci of the ellipse are $(\pm ae,0)=(\pm \sqrt{7},0)$

The radius of the required circle = $\sqrt{{\left(\sqrt{7}-0\right)}^2+{\left(0-3\right)}^2}=\sqrt{7+9}=\sqrt{16}=4$

Hence, 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 $(0,3)$ is $4$.