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

Foci are given by $(\pm ae,0)$

As $a^2{e}^2=a^2-b^2=7$ we have equation of circle as

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

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