Answer
The path is parametrized by
${\bf{r}}\left( t \right) = \left( {2 + 4\cos 2t,10, - 3 + 4\sin 2t} \right)$
Work Step by Step
A circle in the $xz$-plane with radius $4$ and centered at $\left( {x,z} \right) = \left( {2, - 3} \right)$ is given by
${\left( {x - 2} \right)^2} + {\left( {z + 3} \right)^2} = 16$
Since parametrization is not unique, it can be parametrized by $x\left( t \right) = 2 + 4\cos \alpha t$ and $z\left( t \right) = - 3 + 4\sin \alpha t$, where $\alpha$ is a scalar; because it satisfies the equation above.
A circle in the plane $y=10$ with radius $4$ and center $\left( {2,10, - 3} \right)$ can be parametrized by
${\bf{r}}\left( t \right) = \left( {2 + 4\cos \alpha t,10, - 3 + 4\sin \alpha t} \right)$,
where $\alpha$ is to be determined.
The derivative is
${\bf{r}}'\left( t \right) = \left( { - 4\alpha \sin \alpha t,0,4\alpha \cos \alpha t} \right)$
Since the speed is $8$, so
$||{\bf{r}}'\left( t \right)|| = \sqrt {16{\alpha ^2}{{\sin }^2}\alpha t + 16{\alpha ^2}{{\cos }^2}\alpha t} = 8$
$4\alpha = 8$, ${\ \ \ }$ $\alpha = 2$.
Therefore, the path is parametrized by
${\bf{r}}\left( t \right) = \left( {2 + 4\cos 2t,10, - 3 + 4\sin 2t} \right)$
