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

Answer 1

Answer: (B)

$4$

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

Comparing it to the standard equation for ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ we get

$\Rightarrow a=4,b=3$

Now the coordinates of foci are given by $f:(\pm \sqrt{a^2-b^2},0)$

Substitute values of $a$ and $b$ we get coordinates of $f$ to be $f:(\pm \sqrt{7},0)$

Now the center of the circle is given to be at $(0,3)$

Since foci lies on the circle, to find the radius we can find the distance between the foci and the center.

Radius, $r=\sqrt{{\left(\sqrt{7}\right)}^2+3^2}=4$