Ellipses

We have a circle with centre $(3,0)$ :

$(x-3)^{2}+y^{2}=r^{2}$

In this case, the ellipse is vertically more stretched out than horizontally. So the foci must have their x-coordinates being equal to 0.

The distance (c) between a focus and the centre of the ellipse is:

$c = \sqrt{\sqrt{16}^{2}-\sqrt{9}^{2}} = \sqrt{7}$

So, the coordinates of the upper focus is $(0,\sqrt{7})$ .

Putting these coordinates into the equation of the circle, we get:

$(0-3)^{2}+\sqrt{7}^{2}=16=r^{2}$

So, the equation of the circle, passing through the foci of the given ellipse:

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

Rewriting this equation:

$\boxed{x^{2}+y^{2}-6x-7=0}$