Hi,
Heating of 7805 is common thing but as a system designer, we must understand the concept explained below.
I am assuming that your circuit is consuming 500mA urrent during operation.
Case-1: When you are connecting 12V adaptor, you will get 5V output. Do you know where the remaining 7V goes?
We will get 7V drop accros IC 7805. You can measure this drop accross pin 1 and 3. When circuit will consume 500mA current, heat disipation accross IC will be 7V*500mA= 3.5W
Case-2: When you are connecting 9V battery, you will get 5V output. Do you know where the remaining 4V goes?
We will get 4V drop accros IC 7805. You can measure this drop accross pin 1 and 3. When circuit will consume 500mA current, heat disipation accross IC will be 4V*500mA= 2W
In both case your requrements of 5V and 500mA current is fulfilled but still you are wasting 1.5W extra in first case because of wrong selection of power supply. this concept is very much essential while designing real world applications specially battery operated device.