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

Answer 1

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

All given equation of circle in option have same center at $(0,3).$

Now,

eccentricity of ellipse,e $=\sqrt{1-\frac{b^2}{a^2}}$

e $=\sqrt{1-\left(9/16\right)}=\sqrt{7}/4$

$\Rightarrow$ foci of ellipse $=(\pm ae,0)=(4\times \sqrt{7}/4,0)=(\pm \sqrt{7},0)$

$(\sqrt{7}-0{)}^2+(0-3{)}^2$

The center of the circle is $(0,3)$

radius $=\sqrt{(\sqrt{7}-0{)}^2+(0-3{)}^2}$$=\sqrt{7+9}=4$.

Hence, the equation of the circle is :

$(x-0{)}^2+(y-3{)}^2=4^2=16$

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