Answer
(a) $x^2+y^2=4$
(b) See explanations below.
Work Step by Step
(a) Divide 16 on both sides of the ellipse equation, we get $\frac{x^2}{4^2}+\frac{y^2}{2^2}=1$, Compare it with a standard ellipse equation, we get $a=4, b=2$. Thus the equation for the ancillary circle can be written as
$x^2+y^2=4$ which is centered at $(0,0)$ with radius $r=b=2$
(b) Given point $(s,t)$ on the ancillary circle, we have $s^2+t^2=4$. To test if point $(2s,t)$ is on the ellipse, plug the coordinates into the elliptical equation to get $(2s)^2+4y^2=4(s^2+t^2)=4\times4=16$ indicating that the point is indeed on the ellipse.
