Prove that as $PP'$ varies,the circle generates the surface $(x^2+y^2+z^2)(\frac{x^2}{a^2}+\frac{y^2}{b^2})=x^2+y^2.$ $POP'$ is a variable diameter of the ellipse $z=0,\frac{x^2}{a^2}+\frac{y^2}{b^2}=1,$ and a circle is described in the plane $PP'ZZ'$ on $PP'$ as diameter.Prove that as $PP'$ varies,the circle generates the surface $(x^2+y^2+z^2)(\frac{x^2}{a^2}+\frac{y^2}{b^2})=x^2+y^2.$

My Attempt
Let the circle be described on the plane $PP'ZZ'$.Let $P$ be $(a\cos\theta,b\sin\theta)$ and $P'$ be $(-a\cos\theta,-b\sin\theta)$.
Then the equation of the circle be $(x-a\cos\theta)(x+a\cos\theta)+(y+b\sin\theta)(y-b\sin\theta)=0,z=0$
$x^2-a^2\cos^2\theta+y^2-b^2\sin^2\theta=0,z=0$
I am stuck here.Please help.

The circle has two endpoints $(-X,-Y)$ and $(X,Y)$ with $\dfrac{X^2}{a^2}+\dfrac{Y^2}{b^2}=1$ and is perpendicular to the $z$-plane.
Let $(x,y,z)$ be a point on this circle.
Then we have:
$x^2+y^2+z^2=X^2+Y^2$ and $y/x=Y/X$.
So let $Y/y=X/x=c$.
We get: $c^2\left (\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2} \right)=1$
as well as: $x^2+y^2+z^2=c^2(x^2+y^2)$. Multiplying these two equations gives the desired equation.
