Coordinates, and Earth-based telescopes should be able to do it.
Snarfed from another web site:
Quite simple, really.Size of Lunar Module. Let's be really generous and say 10m square. Distance between Hubble and Moon. About 350, 000km. This works out as an visual angle of (10m)/(3.5 x 10^8m) * (180/PI) = 1.6 x 10^-6 degrees = 6 milliarcseconds. The WFPC2 'telescope' on Hubble has the following resolution: 800x800 pixels of a 35 arcseconds field of view with a pixel scale of 46 milliarcseconds. Actually resolution in practice is a little below this.So what does this all mean? Well, roughly speaking, it means that the LM would have to be 15 times larger before it would even cause a dot on a Hubble picture.
Heard someplace that the Hubble's tracking is too slow. The Moon is too fast. The few Hubble pictures of the Moon were
just point where its going to be and start snapping, as it were.
So, to try to get even a specific 100m2 area...