Why does a virtual ground form for differential pair?
A virtual ground is a net that isn't connected to ground but behaves (under proper circumstances) as if it is. A common example is this circuit:



The opamp keeps the voltage at the Virtual ground node 0 V (same as the opamp's + input) but the net isn't connected to ground.

Something similar happens in the differential pair. When we apply a differential signal then the voltage on the common-source net does not change. Let's say that \$V_{G1}\$ increases 1 mV while at the same time \$V_{G2}\$ decreases 1 mV.

So \$V_{G1}\$ => + 1 mV, \$V_{G2}\$ => -1 mV

Then in the linearized (small signal) model of the differential pair the voltage on the common source point does not change.

As long as the increase of \$V_{G1}\$ is as large as the decrease of \$V_{G2}\$ (the input signal is differential) this will be the case.

The symmetry is important because if for example Q1 had a larger W/L than Q2, then the change of \$V_{G1}\$ would have more impact at the Q1 side compared to the Q2 side (because Q1 is stronger). That would effect the voltage at the common source. Q1 would "win" and the voltage at the source would go up instead of remain the same (Q1 pulls "harder" up than Q2 can pull down).