Intersection point of two circles in $3$-D space I have two circles in $3$-D space both centered at the origin $(0,0,0)$, and both the same radius $r$. 
In my coordinate system $x$ is "right", $y$ is "up", $z$ is "forwards".
One is formed by taking a circle in the $y-z$ plane then rotating it $\theta$ around the $y$ axis. The other is formed by taking a circle in the $x-z$ plane and rotating it $\phi$ around the $x$ axis. 
I want to know the points at which the circles intersect for arbitrary $\theta$ and $\phi$.
Thank you.
 A: Adding to the comment of Ivan, you can begin by finding the equation of the two planes that contain the two circles. $(p_1):A_1x+Γ_1z=0, (p_2):B_2y+Γ_2z=0$
The $yz-plane$ is given by the equation $x=0 \implies$ the normal vector $ n=(1,0,0)$
You can now find the equation of $(p_1)$ by rotating $n$ by the angle $θ$ to give you $n_1=(A_1,B_1,Γ_1)$. You can do this directly or you can use the $3 \times 3$ rotation matrix i.e.:
$$
        \begin{pmatrix}
        cosθ & 0 & \pm sinθ \\
        0 & 1 & 0 \\
        \mp sinθ & 0 & cosθ \\
        \end{pmatrix}
        \begin{pmatrix}
        1\\
        0\\
        0\\
        \end{pmatrix}
$$
and set the sines are according to the way of rotation.
Similarly for the second circle.
Continue by solving the system of $(p_1),(p_2)$ to find the line of intersection $(l):x=x_1+ta_1,y=y_1+ta_2,z=z_1+ta_3, t \in \Bbb R$ and solve the system $(l),(s)$, where $(s):x^2+y^2+z^2=r^2$ to find $t$. Now substitute $t_1,t_2$ back into $(l)$ to find the the coordinates of the two points.
