How long was the exposure? Can you show us the shot?
You dou know that exposures of more than a few seconds will cause star trails, right?
Without a tracking head, the best option in your lens lineup is the 10mm.
You have to focus manually and it is difficult and often involves a bit of trial and error.
You'll have to invest the time to focus before shooting and it may take 10 minutes or so to refine focus to a point where you're happy... focus as close as possible then take a test shot (I crank up the ISO to max to decrease the exposure time for these test-shots) then very very gently tweak focus (remember which direction you adjusted) and take another... then compare to find out if you're getting better or worse. Keep working it until you're happy (again... this will take a while.)
But those 10 minutes or so of time investment on refining focus will be rewarded ... return the camera exposure to more sane settings (so you're no longer shooting at max ISO) and start taking the shots you want -- confident that you're images will be nicely focused.
Since the Earth is spinning (we rotate from west to east at a rate of 15 arc-seconds of angular rotation per second of real time) the amount of movement for any given star depends on angle of view for your lens/camera combination.
For a full-frame camera, the simple math is to divide "600" by the focal length of the lens, but if you want to be conservative (because at 600 you may see a tiny bit of elongation on the stars) divide "500" by the focal length of the lens.
E.g. if you had a full frame camera (the 70D is not full frame) and, say, a 14mm lens then it would be 600 ÷ 14 = 42.9 seconds (about 43 seconds).
But to be conservative about the star elongation you might really do 500 ÷ 14 = 35.7 (about 36 seconds).
But the 70D isn't a full-frame camera. It has an APS-C sensor with a 1.6x crop factor. So we have to divide the 600 (or 500 if being conservative) by 1.6. 600 ÷ 1.6 = 375. Suppose you use your 10mm lens ... that would be 375 ÷ 10 = 37.5 seconds.
To be conservative, it would be 500 ÷ 1.6 = 312.5 ... and divide that by 10mm and you'd get 31.3 seconds.
Youi can go much longer IF your camera is mounted to a tracking head. You can also use longer focal lengths and get much more detail in the sky. Here's a shot I captured using a tracking head... and a 135mm lens.
Without a tracking head, the longest shot one could reasonable catpure on an APS-C body (this was captured using a Canon EOS 60Da body) would be 312.5 ÷ 135 = 2.3 seconds... not nearly long enough. But with a tracking head the shot can be many minutes long (notice there is no Earth landscape in this image - it's all sky).
The market has evolved a bit since I first started offering advice on this and now the two major vendors for tracking heads (the popular models) are:
1) The "Star Adventurer" head by Sky Watcher
2) The "SkyTracker Pro" head by iOptron
Both of these are heads which attach to a photo tripod (preferably a solid photo tripod). The head is tilted to point at the celestial pole (they come with aids which help you align them) and this way the axis of the head rotation is parallel to the axis of Earth's rotation. As the Earth rotates from west to east at 15 arc-seconds per second, the tracking head rotates form east to west at 15 arc-seconds per second... cancelling out the rotation of the Earth and allowing you to take very long exposures of the sky with tack-sharp results. (I've done 8-minute exposures without the tiniest hint of star elongation).
That 15 arc-seconds per second rotation speed is also called "sidereal" rate (pronounced "sid EAR ee al"). "sidereal" means "of the stars". The greek word for iron was "sidero". And since meterorites "appear" to be the remains of falling stars (we now that's not really what they are, but it is a long-held belief) and are mostly made of iron, they assumed that iron came from the stars (and that part is true, but the process by which it happens is a bit different then they understood). So anything related to the stars was called "sidereal". "Sidereal rate" is the rate at which the stars appear to move across the sky.
If you want landscape in the image, then the stars will be sharp and the land will be blurred (because the head is rotating). But I think both of these heads have a 1/2 sidereal speed. This is a great compromise speed because while it's not fast enough to keep up with the motion of the stars for any length of time, it does DOUBLE the amount of time you can image before the stars appear to elongate. So if you were previously only able to image for 30 seconds, now you can image for 60. And the land will appear to twist... but that same rule applies... you can go the same length of time before the landscape appears to blur.
