First, depends on the narrowest section of the scope that the light path travels. Usually this is at the focuser, or diagonal if you use one. Secondly, focal reducer will shrink this down.
Roughly speaking, if you were to image through a 2" focuser, you would have an image circle (possibly with field curvature and some vignetting) towards the edges at a bit under two inches diameter. If you put a focal reducer in the path (such as the Meade .63 reducer you used) then the clear aperture of that reducer is 1.5" (if you look at the reducer 2" is the outer diameter of the metal casing, 1.5" is the clear aperture of the glass), and ontop of that it reduces that 1.5" down by approx 0.63 of that, so 0.945". If you use a 0.33 reducer you get half of this again, less than half an inch, which is why you can only use small CCD chips with these stronger reducers.
If you want to image on full 6cmx7cm medium format film, you will need a much larger focuser, which is why high end apos have 2.7" and 4" focusers and massive field flatteners.
I think this is what you are asking (based on what you asked me about vignetting the other day
) If not, image circle is based on the focal length of the scope and the size of the image sensor (or film) and someone else can give the mathematical answer
