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)

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

Coordinate of foci are $(\pm ae,0)$

$a=4,b=3,e=\sqrt{1-\frac{9}{16}}=\frac{\sqrt{7}}{4}$

Radius of circle$=\sqrt{(ae{)}^2+3^2}$

$=\sqrt{16}=4$

Equation of circle

$(x-0{)}^2+(y-3{)}^2=16$

$\Rightarrow x^2+y^2-6y-7=0$.