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$$\begin{align*} x &= a \sin \theta \cos \varphi\\ y &= a \sin \theta \sin \varphi\\ z &= a \cos \theta \end{align*}$$

Given this is an equation of a sphere, how would I find its center and radius?

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If you want to get an intuitive picture of what is going on with these equations, you might like to consider the maximum and minimum values of the $x, y, z$ coordinates.

Since the trig functions vary between $+1$, and $-1$, it is easy to see that the maximum and minimum values of $x, y, z$ are $a$ and $-a$, so the centre is halfway between the maximum and minimum i.e. at $(0,0,0)$ and then the radius is clear.

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It is centered at $(x,y,z)=(0,0,0)$ and its radius $= a$

You can do this by inspection of the conversion between Cartesian and Spherical Coordinates:

$$\begin{align*} x &= r \cos\phi \sin\theta\\ y &= r \sin\phi \sin\theta\\ z &= r\cos\theta \end{align*}$$

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Please use \cos and \sin. – Did May 23 '12 at 18:04
Can you be a bit more detailed? Why is the radisu = a? – Tool May 23 '12 at 18:10
Compare the problem you wrote with what I wrote symbol for symbol. If you replace r with a in mine you get yours. Make sense? – jds May 23 '12 at 18:22

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