Question

The equation of the circle passing through the foci 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

Answer: (A)

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

$a=4,b=3,e=\sqrt{1-\frac{9}{16}}\Rightarrow \frac{\sqrt{7}}{4}$
$\therefore$ Foci $=(\pm ae,0)=(\pm 4\times \frac{\sqrt{7}}{4},0)=(\pm \sqrt{7},0)$
Radius of the circle, $r=\sqrt{(ae{)}^2+b^2}$
$=\sqrt{7+9}=\sqrt{1}6=4$
Now, equation of circle is
$(x-0{)}^2+(y-3{)}^2=16$
$\therefore x^2+y^2-6y-7=0$