Finding points relating to the edge of a circle in an x,y coordinate system

My question is a bit hard for me to express, so please bear with me. I never got far in trig, and haven't done much on the subject in years; trying to get back into it as it's a pretty major part of physics and I'm starting game development as a hobby.

Basically, I want to draw a pattern in the centre of a screen. That pattern will consist of a random amount of lines, all the same random length. These lines will all originate from one centre point, and will all be evenly spaced.

I'm working with a grid of pixels. Anyone familiar with graphical programming will know that most pixel grids start with the top left 0,0 in an x,y co-ordinate system. There are no negative points. Say my grid is 500x500. The centre pixel will be located at 250,250. 0,0 is the top-left pixel, 500,500 is the bottom right. All measurements in this question are in pixels.

Now say that this centre pixel, 250,250, is the exact centre of a circle. From now on let's call this centre point xy. The radius of the circle is r (assume it fits inside the grid).

There is another variable in play. Let's call it a. a is the amount of lines I want drawn from the centre of the circle to the edge of the circle, all evenly spaced (so r is the length of each line). I'm assuming that in order to do this, I have to do this: 360 / a, which gives me a degree (I'm totally making these variable names up) which I will somehow use to evenly space the lines. This leads to the question!

Given xy, r & a, how can I go about getting the x & y co-ordinates of the lines' end-points?

I hope this is clear enough! Please let me know if I need to clarify anything.

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Each line extends from $(x,y)$ to $(x+r \cos \theta, y+r \sin \theta)$. You are right that if you want $a$ lines to fill up the circle, they should be spaced at $\frac {360}a$ degree intervals. So use $\theta = \frac {360n}a$ for $n=0,1,2,\ldots a-1$
@Djentleman: You want a total of $a$ lines. I was presuming you start with one at $\theta = 0$, so if you count there will be $a$ of them. The next one would be $\frac {360a}a=360$ and lie on top of the first. This equally spaces the lines in $\theta$. Is that what you want? –  Ross Millikan May 28 '12 at 10:46
@Djentleman: Are your pixels the same size in each direction? If your grid is $500 \times 500$ but the window is not square you need to correct for that. –  Ross Millikan May 28 '12 at 11:13