Question

The equation of the circle passing through the foci of the ellipse $\frac{x^2}{9}+\frac{y^2}{16}=1$ and having the centre at (0, 3) is

(IIT JEE Main 2013)

• 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

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

We have, equation of ellipse as $\frac{x^2}{9}+\frac{y^2}{16}=1.$
Here, a = 4, b = 3
So, $e=\sqrt{1-\frac{9}{16}}=\frac{\sqrt{7}}{4}$
Foci of the ellipse are:$(\pm ae,0)=(\pm 4\times \frac{\sqrt{7}}{4},0)=(\pm \sqrt{7},0)$
Radius of the circle is the distance between the centre of the circle and point on the cirlce.
So, $r=\sqrt{(\sqrt{7}{)}^2+3^2}=\sqrt{7+9}=\sqrt{16}=4$
So, the equation of the circle is
$(x-0{)}^2+(y-3{)}^2=16$
$\Rightarrow x^2+y^2-6y-7=0$