Since its pretty hard to take screenshots of a camera system, you lucky people get some derivations accompanied by shitty MSPaint diagrams (well, only one shitty MSPaint diagram, but when you've seen one, you've seen them all). Enjoy!
The whole idea with spherical coordinates is that you map cartesian coords to the surfaces of a sphere and vise versa. The camera sits at the center of the sphere, and looks out at the point on the surface. This requires a little bit of trig, but nothing especially complicated.
Points in spherical coords are not represented with (x, y, z) values, but rather (r, ?, ?). r is the sphere's radius, ? is your angle from the x axis, and ? is your angle from the z axis (usings D3D's left-handed coordinate system):
Basically, all I need to do for my camera is represent a target in spherical coordinates. When its time to draw, I can convert this point to cartesian coordinates, which will in effect give me a vector from my camera to the target. The conversian goes something like this:
sin(?) = y / r -> y = rsin(?).
cos(?) = r' / r -> r' = rcos(?).
sin(?) = z / r' -> z = r'sin(?) -> z = rcos(?)sin(?).
cos(?) = x / r' -> x = r'cos(?) -> x = rcos(?)cos(?).
So, your cartesian point P = (x, y, z) is:
(rcos(?)cos(?), rsin(?), rcos(?)sin(?)), where ? / 2 > ? > -? / 2 and 2? >= ? >= 0.
By keeping a cartesian point Q = (x, y, z) that represents my camera's location in world space, I can get the 'at' vector for the lookat matrix simply by adding Q + P (Note that, even though P and Q are points, it is possible to simply treat them as vectors).
Moving the camera parrallel to this 'at' vector is easy: just scale 'at' by however much you want to move and add or subtract it to Q to move forward or backward. Moving perpendicular requires one extra step, but its not hard. The cross product of 'at' with your global 'up' vector (I use (0, 1, 0) in D3D) will give you the necessary perpendicular vector (but be carefull, this camera will break if you allow it to look straight up or down). Now, just scale and add/subtract that perpendicular vector to Q and you're all set.
Not exactly groundbreaking stuff, but I thought it was pretty nifty, and hopefully you got a little something out of this post too.