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

For the ellipse $\frac{x^2}{16}+\frac{y^2}{9}=1$, we have $a^2=16,b^2=9$ Let e be the eccentricity of the ellipse. Then

$e=\sqrt{1-\frac{b^2}{a^2}}=\sqrt{1-\frac{9}{16}}=\frac{\sqrt{7}}{4}$

The coordinates of the foci are $(\pm ae,0)=(\pm \sqrt{7},0)$

$\therefore$ Radius of the circle = Distance between $(\pm \sqrt{7},0)$ and ( 0, 3)

$=\sqrt{7+9}=4$