Parametrizing all circles of circular cone by intersecting it with sphere of radius $r>0$

Question

Let's assume that we have a downwards circular cone whose angle of inclination is $\theta$, and a sphere of radius $r>0$, both of which are centered at $(0,0,0)$. Then, how could we parametrize all circles of circular cone by intersecting it with the sphere? In the diagram, we exclude the yellow, elliptic cone.

Our sphere is $\mathbb{S}_r^1=\{(x,y,z)\in \mathbb{R}^3:x^2+y^2+z^2=r^2\}$, and our cone is $\displaystyle C=\left \{(x,y,z)\in \mathbb{R}^3:z=\frac{1}{\tan \theta}\sqrt{x^2+y^2}\right \}$. Then, $\displaystyle x^2+y^2+z^2=\left (\frac{1}{\tan ^2\theta}+1\right )(x^2+y^2)=r^2$, and so, $\displaystyle y=\pm \sqrt{\frac{r^2-\frac{1}{\tan ^2\theta}x^2}{\frac{1}{\tan ^2\theta}}}$.

Answer 1

(see figure below).

Right triangle $OAB$ results from a vertical section by a plane containing the $z$ axis.

Pythagoras relationship gives :

$$\rho^2=r^2-z_0^2$$

Therefore, the generic circle situated at altitude $z=z_0$ has the following equations :

$$\begin{cases}x^2+y^2&=&r^2-z_0^2\\z&=&z_0\end{cases}$$

or, under the form of 3 parametric equations:

$$\begin{cases}x&=&\sqrt{r^2-z_0^2}\cos \theta\\y&=&\sqrt{r^2-z_0^2}\sin \theta\\z&=&z_0\end{cases}$$