Here's a really (and I mean really!) quick and dirty set of calculations that might give you an idea of the magnitudes of settlement you could be dealing with.
The settlement potential of the tank location can be determined a number of ways, but probably the best thing to do would be a plate load bearing test. The test can be run to simulate the range (though not the duration) of loads you are expecting. A test like this will give you a spring constant $k$ that represents the "modulus of subgrade reaction" of the bearing soil (for the tested loading range). However, it's a short term test that doesn't take into account creep, so the long-term $k$ value will be lower.
In general, a short-term $k$ will run from something like 80pci for a very soft clay to something like 250pci for a very dense sand (caveat: this is just from the top of my head without looking anything up).
So let's use the worst case scenario here, and to take into account creep, let's do what geotech engineers do best and slap a 2.5 safety factor on it. So we have about a 30pci modulus of subgrade reaction.
Let's also assume that most of the differential settlement will occur as a result of the uneven loading of the empty tank, and that the emptying/filling of the tank is going to have a negligible contribution to the differential settlement. This isn't too terrible of an assumption, since the difference in applied surface pressure (which determines differential settlement) is much greater in the empty state, and it's also conservative because it will only be empty 10% of the time anyway.
So here we go (I'm American, so we're doing everything other than what you gave for dimensions in imperial-scum units first and then converting - sorry!):
$k = 30 \frac{lbf}{in^3}$, $\gamma_{concrete}=150\frac{lbf}{ft^3}$, $H_{concrete}=2m$
Applied pressure under half the tank: $q_c=H_c\times\gamma_c=0.98ksf=5.3 \frac{tonf}{m^2}$
Settlement under loaded half of tank: $S=\frac{q_c}{k} = 0.23in = 5.8mm$
If we assume the other side of the tank does not settle at all, our differential settlement comes out to about 6mm.
Now, this number assumes the loaded side of the tank is free to settle while the unloaded side remains static. This is not the case. Assuming the tank is nice and stiff, some of the applied pressure on the loaded side will be transferred to the unloaded side (which will reduce the settlement of the loaded side).
I don't know what the application is for this tank, but the above is probably a pretty conservative analysis of the situation you described. I would be surprised if differential settlement potential turns out to be a problem for you.
EDIT: One thing to note is that the tank will "wiggle" when it is being filled/drained. What I mean is, the entire thing will settle more when it is filled, but it will settle more in the unloaded side (thereby undoing some of the differential settlement in the empty condition). Then when drained, the soil will rebound and the tank will return to the more-tilted empty condition when the unloaded side rebounds more than the loaded side (though it is likely neither side would rebound fully).
Assuming the 6mm of settlement from above, the deflection angle for the 24m diameter tank comes out to be $\arctan\frac{6mm}{24m}=0.014^{\circ}$. Pretty tiny.
