Any hints as to how I can find the equation of all great circles passing through a given point (polar angle $\theta$, azimuthal angle $\phi$) on the surface of a unit sphere? Thanks.


How do you want to index all of the possible great circles?

For this answer I assume that we index them by the tangent vector, because given $(x_0, y_0, z_0)$ and a vector $\vec{u}$ in the tangent plane to the circle at $(x_0,y_0,z_0)$, there is a great circle going through $(x_0,y_0,z_0)$ whose tangent vector is $\vec{u}$. Note that the set of tangent vectors is not in one to one correspondence with the set of great circles--in order to have that we could have to only consider vectors of unit length and then consider vectors which are negative of each other as equivalent.

Now we find an equation of the unit circle through $(x_0, y_0, z_0)$ in the direction of $\vec{u}$. The plane containing the vectors $<x_0,y_0,z_0>$ and $\vec{u}$, when intersected with the sphere creates the great circle we want. This is because $<x_0, y_0, z_0>$ is normal to the sphere at $(x_0, y_0, z_0)$. This plane has the equation $$(\vec{u} \times <x_0, y_0, z_0>) \bullet <x,y,z> =0.$$ So that $$(\vec{u} \times <x_0, y_0, z_0>)_x \cdot x + (\vec{u} \times <x_0, y_0, z_0>)_y \cdot y + (\vec{u} \times <x_0, y_0, z_0>)_z \cdot z =0.$$

We can solve for $z$ (assuming the coefficient on $z$ is nonzero) and substitute back in to $x^2 + y^2 + z^2 =1$. $$x^2 + y^2 + \left(\frac{(\vec{u} \times <x_0, y_0, z_0>)_x \cdot x + (\vec{u} \times <x_0, y_0, z_0>)_y \cdot y}{(\vec{u} \times <x_0, y_0, z_0>)_z}\right)^2 =1$$ Which is an implicit equation for the circle that you seek. You can convert to spherical afterwards if you like. I did it this way so I could work with planes in cartesian coordinates instead of spherical.

Here is a little image I created of the setup, if my words were not clear. The point $(x_0, y_0, z_0)$ is the black point. The green plane is the tangent plane. $\vec{u}$ is the red arrow. The pink plane is the one whose intersection with the sphere carves out a great circle.


  • $\begingroup$ Thanks. Excuse my hopeless vector algebra but how would that work with, for example, a point on the “north pole” $<x_{0},y_{0},z_{0}>=<0,0,1>$ and with a tangent vector $\vec{u}=<1,0,0>$ along the $x$ axis, which should give a great circle on the $xz$ plane? I can't see how that great circle pops out of your equation. $\endgroup$ – Peter4075 Sep 28 '14 at 16:15
  • $\begingroup$ Find the equation of the plane containing those two vectors. In this case it is $y=0$. Plugging $y=0$ into $x^2+y^2+z^2=1$, we get $x^2+z^2=1$. This is the equation of the great circle. $\endgroup$ – Tom Hallward Sep 28 '14 at 20:32
  • $\begingroup$ Bravo! The final equation in your excellent answer confused me. I've just worked out the equation for a point on the equator with a tangent vector along the $x$ axis and out pops the equation for the equatorial great circle. $\endgroup$ – Peter4075 Sep 29 '14 at 5:45

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