q point keeps shifting when there is input signal
I think that the correct Q-point is 6.5V, not 5.5V.  The Q-point should be roughly halfway between Vcc and the DC emitter voltage.  Halfway between 12V and 1V is 6.5V, not 5.5V.

At 5.5V, the negative swing is just 5.5V - 1V = 4.5V while the positive swing is 12V - 5.5V = 6.5V.  This is why the negative peak clips first.

UPDATE:  I wrote the above under the assumption that, as in the previous circuits the OP has asked about, the emitter resistor is AC bypassed.  Now that the schematic has been posted, I see that there is no emitter bypass capacitor in this circuit.

With the emitter resistor bypassed, the emitter is at AC ground and this simplifies the Q-point calculation.  Without bypass, the calculation is more complex.

When the transistor is saturated, the voltage between collector and emitter is roughly zero.  Thus, the emitter current is:

\$i_{E(max)} = \dfrac{V_{CC}}{R_5 + R_7} =  17mA\$

Thus, the lowest voltage the collector can fall to is:

\$v_{C(min)} = i_{E(max)} \cdot R_7 = 1.56V \$

So, a rough estimate of the Q-point collector voltage is:

\$(12 - 1.56)/2 + 1.56 = 6.8V \$

However, this is an iterative process.  When you change the value of the collector or emitter resistor, the above calculations change so the idea is to calculate a reasonable starting point, simulate to see where you are, and adjust / tweak as necessary.

UPDATE 2:  For a detailed analysis of the calculation of clipping levels, see Single Stage BJT Amplifier Clipping Levels by W. M. Leach.