What would be the waveform of output of this circuit?
Because of C1 the DC on the left side can be ignored. It will make no difference1 if the 5 V DC supply is there or not.

1 If there is DC present then C1 must be rated for the DC voltage and if it is a polarised capacitor such as an electrolytic it must be inserted in the correct orientation.


Without D1 the output voltage will be a sine wave with the same amplitude as the input but swinging positive and negative.
With D1 inserted the output will be "clamped" or limited to 0 V if it tries to go negative. Your question shows that you understand that there will be a 0.7 V drop across the diode if significant current is flowing. Therefore Vo will be clamped to -0.7 V when the input goes negative.
Once we go past the bottom of the input sinewave the input voltage is increasing from \$ V_{pk-} \$ to \$ V_{pk+} \$. The right side of C1 will rise with this and so will go to \$ V_{pk+} - V_{pk-} - 0.7\ \text V = 2V_{pk} - 0.7 \ \text V\$. Note that the 5 V makes no difference.




Figure 1. Running your circuit with Vi = 5 V AC on top of 5 V DC we get the results above.


The left side of C1 reaches +10 V.
The right side of C1 reaches +5 V (because that is how much AC we have seen up until now).
At point three the left side of C1 is back where it started.
After this point the AC is clamped at -0.7 V and so it will rise to \$ 2V_{pk} - 0.7 \ \text V \$.


Your English is fine!