Circle On Sphere 
Sorry if this sounds too silly but  my math skills are very poor and I just need this problem fixed. 
I made this graphic with geogebra 3D and it was quite easy there but I don't know how to write the equations for this.
In the image you can see:


*

*Sphere with radius 1. 

*Point $B$ on the sphere with spherical coordinates.

*Point $C$ on the sphere with spherical coordinates.

*$\alpha$ is the angle between $B$ and $C$.

*Circle $d$ on the sphere from segment $b$ through point $C$.

*$G$ and $D$ are points on that circle.

*$\theta$ is the angle between $C$ and $D$ from the center of the circle.


What I want is an equation that, given and Initial Point ($B$), and $\alpha$ angle, gives me points on the circle $d$ ( like $C$, $G$ or $D$) with $\theta$ as a parameter.
And if possible in spherical coordinates... 
Thanks 
 A: Construct a parametrization of the circle $d$ in $3$ stages of increasing generality.
1.  Suppose that $B$ is the north pole (colatitude $\varphi = 0$).
The circle is the parallel (line of constant colatitude $\varphi = \alpha$), which has $z = \cos \alpha$ and $r = \sin \alpha$; hence, it is parametrized in rectangular coordinates by $t \mapsto \big( x(t), y(t), z(t) \big)$, where
$$
\left\{
\begin{align}
x &= \sin \alpha \cos t \\
y &= \sin \alpha \sin t \\
z &= \cos \alpha
\end{align}
\right.
\qquad \text{for } 0 \le t < 2\pi.
$$
2.  Suppose that $B$ is on the prime meridian (longitude $\theta = 0$) but colatitude has some value $\varphi = \beta$, where $0 < \beta \le \pi$.
We take the coordinates $(x, y, z)$ of the circle with center at the north pole (from 1.) and rotate them through an angle $\beta$ along the great circle that includes the prime meridian.  Using standard formulas for rotating coordinates, we have
$$
\begin{bmatrix}
x \\ y \\ z
\end{bmatrix}
\mapsto
\begin{bmatrix}
x \cos \beta + z \sin \beta \\ y \\ -x \sin \beta + z \cos \beta
\end{bmatrix}.
$$
3.  Suppose that $B$ is anywhere on the sphere with spherical coordinates $(\varphi, \theta) = (\beta, \gamma)$.  Take the resulting coordinates from 2. and rotate them about the polar axis of the sphere through an angle of $\gamma$:
$$
\begin{bmatrix}
x \\ y \\ z
\end{bmatrix}
\mapsto
\begin{bmatrix}
x \cos \gamma - y \sin \gamma \\ x \sin \gamma + y \cos \gamma \\ z
\end{bmatrix}.
$$
Putting these together, the general case for a circle $d$ with center at $(\varphi, \theta) = (\beta, \gamma)$ and central angle $\alpha$ (that determines the radius) has rectangular coordinates:
$$
\left\{
\begin{align}
x &= \phantom{-}( \sin \alpha \cos \beta \cos \gamma ) \cos t + ( \sin \alpha \sin \gamma ) \sin t - (\cos \alpha \sin \beta \cos \gamma ) \\
y &= -( \sin \alpha \cos \beta \sin \gamma ) \cos t + ( \sin \alpha \cos \gamma ) \sin t + (\cos \alpha \sin \beta \sin \gamma ) \\
z &= \phantom{-}( \sin \alpha \sin \beta ) \cos t + \cos \alpha \cos \beta.
\end{align}
\right.
$$
A: By my math, the final result should be:
$$x = (\sinα\cosβ\cosγ)\cos t - (\sinα\sinγ)\sin t + (\cosα\sinβ\cosγ)$$
$$y =(\sinα\cosβ\sinγ)\cos t + (\sinα\cosγ)\sin t + (\cosα\sinβ\sinγ)$$
$$z = -(\sinα\sinβ)\cos t+\cosα\cosβ.$$
