Question

The equation of the circle passing through the foci of the ellipse $\left(\frac{x^2}{16}\right)+\left(\frac{y^2}{9}\right)=1$ and having centre at $\left(0,3\right)$ is

• A. $x^2+y^2-6y-7=0$
• B. $x^2+y^2-6y+7=0$
• C. $x^2+y^2-6y-5=0$
• D. $x^2+y^2-6y+5=0$

Answer 1

The correct option is A

$x^2+y^2-6y-7=0$

Explanation for the correct answer:

Finding the equation of the circle:

Given that, a circle passing through the foci of the ellipse $\left(\frac{x^2}{16}\right)+\left(\frac{y^2}{9}\right)=1$ and having center at $\left(0,3\right)$.

The given equation is of the form, $\left(\frac{x^2}{a^2}\right)+\left(\frac{y^2}{b^2}\right)=1$ and so, $a=4,b=3$

The coordinate of the focii is $\left(\pm ae,0\right)$. Thus calculating the value of $e$

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

Focii of an ellipse is given by $\left(\pm ae,0\right)=\left(\pm \sqrt{7},0\right)$

Radius of the circle will be the distance between one focus $\left(\sqrt{7},0\right)$ and center $\left(0,3\right)$ which is given by

$$\begin{gathered} r=\sqrt{{\left(ae\right)}^2+b^2} \\ r=\sqrt{7+9} \\ r=4\text{units} \end{gathered}$$

Therefore, the equation of a circle having a center $\left(0,3\right)$ and radius $r=4$ is given as

$$\begin{gathered} {\left(x-0\right)}^2+{\left(y-3\right)}^2=16 \\ {x}^2+y^2-6y-7=0 \end{gathered}$$

Hence, option (A) is the correct answer.