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

The correct option is A $x^2+y^2-6y-7=0$


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