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)= \Bigg (\pm 4 \times \frac {\sqrt 7}{4},0 \Bigg ) = ( \pm \sqrt 7,\, 0) $
Radius of the circle, $r = \sqrt {(ae)^2+b^2}$
$ = \sqrt {7+9} = \sqrt16=4$
Now, equation of circle is
$ (x - 0)^2 + (y - 3)^2 = 16$
$\therefore x^2 + y^2 - 6y - 7 = 0$