I guess I should have clarified that in any practical implementation, both of these circuits would have a blocking cap from Vcc to ground and possibly even an RFC to provide isolation from the supply. No signal path through the supply line.
You misunderstand me a little, if you feed a current (I) to a simple parallel LC network at its resonant frequency then the current in both L and C will be Q times I.
Where Q is the quality factor of the tuned circuit. (Energy is exchanged between them during each cycle)
Take an example...
For a particular tuned circuit at resonance X
L = X
C = 10ohms
If the Q of the inductor is 100 and the Q of the Capacitor is very high (>>100)
then at resonance the tuned circuit will look like 1k resistor (Q.X
L = Q.X
C = 100x10)
Stick a volt across the tuned circuit an the current will be 1mA but the current through the inductor or capacitor will be 1v/10ohms = 100mA.
Again if you try injecting a small voltage (V) in series with the capacitor you will find the voltage across the tuned circuit is Q.V
My point was that moving C1 as I suggested would reduce the current passing through the supply decoupling by a factor of Q and also reduce the susceptibility to ripple at the resonant frequency by the same factor.
Jim
Oops...
Forgot to add that as well as moving C1 to the positive rail, C3 should also be moved. ie all refer all RF paths to the positive rail.
If you want thee RF ground to be the negative rail, switch to a PNP transistor and swap the rails or use the common collector circuit.