Question

Equation of the circle passing through the focii of the ellipse $\frac{x^2}{16}+\frac{y^2}{9}=1$ and having centre at $(0,3)$ 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 $x^2+y^2-6y-7=0$

Equation of the ellipse is
$\frac{x^2}{16}+\frac{y^2}{9}=1\Rightarrow a=4,b=3e=\sqrt{1-\frac{b^2}{a^2}}\Rightarrow e=\sqrt{1-\frac{9}{16}}\Rightarrow e=\frac{\sqrt{7}}{4}$

Foci are $(\pm ae,0)\equiv (\pm \sqrt{7},0)$
Equation of circle with centre $(0,3)$ and radius $r$ is
$x^2+(y-3{)}^2=r^2\cdots (1)$
This circle passes throught $(\sqrt{7},0)$
$\Rightarrow (\sqrt{7}{)}^2+(0-3{)}^2=r^2\Rightarrow r^2=16$
Now, from equation $\left(1\right)$
$x^2+(y-3{)}^2=16\Rightarrow x^2+y^2-6y-7=0$