..
and: if you use an 7805 (or any other electronics device) --> read it´s datasheet.
It tells you how to operate it. This is reliable information.
Don´t rely on random internet sources. Most of them contain mistakes, are incomplete...
Rely on
* informations from manufacturers
* universities
* schools
* reputable electronics designers
The manufacturer internet page of the device schold be the first place to visit. For newbies as well as for professionals.
There you find the datasheet and additional helpful informations like application notes, design notes, simulation tools..
This is all you need (indeed more than you need) ...
.. only for special applications you may need additional detail informations. (like when you want to use the voltage regulator as a constant current source. So not used the usual way)
****
To the circuit:
It uses a transformer with 9V output. On no load a small 9V transformer may easily output 12V RMS sine.
The rectified bridge voltage thus may go up to (12V x sqrt(2)) - (2x 0.6V) = 17V - 1.2V = about 16V (first part is the peak of a sine. Second part is the drop of two diodes)
If the output voltage is 5V ... then 11V need to be wasted. This (11V x current) is the (peak) power that has to be dissipated as heat.
On a 0.2A load .. I guess realistically you may expect up to 1.8W of heat (average). --> The 7805 wil get hot. Too hot without heat sink. Thus add a heatsink.
Due to loss (heat) and rather heavy transformer .. nowadays one uses switch mode supplies.
Many devices come with with 5V DC plug supplies. Usually there is always such a power supply laying around. (Maybe from a broken device)
There are rare reasons to nowadays use the "mature design" power supply.
Can you tell us the reason why you have chosen it?
Klaus