Consider a point ($p$) rotated $2\pi$ radians around the point $o$ along two dimensions with the distance between the two points being $r$.
The shape that is traced is a circle.
The parametric equation is $(x,y) = (cos(t),sin(t)),$ and the explicit equation is $x^2 + y ^2 = r^2$.
What are the parametric and explicit equations describing the path of a point rotated about another point a certain distance away along 3 dimensions?
Here is an image of the curve. Part of the curve has been cut out so that we can determine its shape without confusion. The part that is cut out is almost linear. It also contains a cusp. The curve probably intersects itself at the cusp. I call it the veselioid.
In $3$D the rotated point ${\bf p} =(x,y,z)$ shall stay on the sphere of radius $r$ centered at the fixed point (${\bf o}=(x_o,y_o,z_o)$) AND on the plane, passing through that point, whose normal is the vector defining the rotation axis (let's call it ${\bf v}=(a,b,c)$): can you visualize that ?
Upon that scheme, you can proceed in various different ways, according to what you know and/or what you want to achieve.
I report below some hints for you to get an orientation
a) Analytic
Since the point has to be on the sphere and on the plane we shall have $$ \left\{ \matrix{ \left( {x - x_{\,o} } \right)^{\,2} + \left( {y - y_{\,o} } \right)^{\,2} + \left( {z - z_{\,o} } \right)^{\,2} = r^{\,2} \hfill \cr a\left( {x - x_{\,o} } \right) + b\left( {y - y_{\,o} } \right) + c\left( {z - z_{\,o} } \right) = 0 \hfill \cr} \right. $$
From here, if needed, you can proceed , again, in different ways.
One way is to get from the 2nd equation one of the terms, whose coefficient is not null and insert that into the 1st. Suppose, e.g., that $c$ is not null, then $$ \left\{ \matrix{ \left( {c^{\,2} + a^{\,2} } \right)\left( {x - x_{\,o} } \right)^{\,2} + 2ab\left( {x - x_{\,o} } \right)\left( {y - y_{\,o} } \right) + \left( {c^{\,2} + b^{\,2} } \right)\left( {y - y_{\,o} } \right)^{\,2} = c^{\,2} r^{\,2} \hfill \cr \left( {z - z_{\,o} } \right) = - {1 \over c}\left( {a\left( {x - x_{\,o} } \right) + b\left( {y - y_{\,o} } \right)} \right) \hfill \cr} \right. $$ You have got the equation of the ellipse projected into the $x,y$ plane, and the "altitude" $z$ over it, i.e. a split into cylindrical coordinates.
If you know abot conics, you can proceed and find the axes of the ellipse and then its polar equation and thus express the whole parametrically vs. the rotation angle as projected on the $x,y$ plane.
a) Vectorial
You can find a orthonormal basis to $\bf v$, i.e. two independent vectors lying on the plane, which are also othogonal between them, and normalized. Then you linearly combine them, with coefficients $r \cos \theta, r \sin \theta$, where $\theta$ is the rotation angle.