# Deriving Parameterization of Great Circle

I am trying to find a spherical parameterization of the great circle path on a sphere of radius R. I parameterize my sphere as follows:

$(R\sin\theta\cos\phi, R\sin\theta\sin\phi, R\cos\theta)$.

I know that the great circle is the intersection of a plane passing through the origin. The equation of such a plane is $ax+by+cz = 0$.

According to this article http://sgovindarajan.wikidot.com/twosphere, by substituting $x,y,z$ into the equation of my plane I should be able to obtain:

$\cot\theta = c_1(\cos(\phi + c_2))$

where $c_1$ and $c_2$ are simply constants. However, I cannot seem to see how the $\sin \phi$ disappears. I began by isolating the $\theta$'s and $\phi$'s. But I cannot seem to arrive at the above result. Is there a trigonometric property I should be using?

Thanks!

I suppose you got to $$\displaystyle{\frac{-c}{\tan(\theta)}=a\,\cos(\phi)+b\,\sin(\phi)}$$
Now this is equivalent to $$\displaystyle{\frac{-c}{\sqrt{a^2+b^2}}\cot(\theta)=\frac{a}{\sqrt{a^2+b^2}}\,cos(\phi)+\frac{b}{\sqrt{a^2+b^2}}\,\sin(\phi)}$$
The two numbers before $$\cos(\phi),\sin(\phi)$$ represent a point on the unit sphere, since its norm is $$1$$.
In other words it we can define a unique $$\phi_0$$ such that $$\begin{cases}\cos\phi_0=\frac{a}{\sqrt{a^2+b^2}}\\\sin\phi_0=\frac{b}{\sqrt{a^2+b^2}}\end{cases}$$
The equation becomes $$\displaystyle{\frac{-c}{\sqrt{a^2+b^2}}\cot(\theta)=\cos(\phi_0)\cos(\phi)+\sin(\phi_0)\sin(\phi)}=\cos(\phi-\phi_0)$$