Remove R3 and put instead diode 1N4148 . The idea is to allow both signal half cycles have about equal way through a PN junction to prevent C1 collecting any DC. That DC would block the way of the signal to the base of the transistor.
The diode must have opposite direction than the BE junction of the transistor. If your transistor really is PNP as drawn the diode should have its cathode connected to transistor's emitter and anode to the base. If it's NPN as written the diode should be the anode downwards.
Your R3 discharges to some degree the DC which tends to cumulate into C1 but it also loads the signal source and cause losses. I guess that's why you found some optimum R3. If your signal source was as drawn (=no series resistance nor other current limitation) the output of the amp would increase as R3 decreases. Finally it would not be an amp at all because dissipation in R3 would be more than the output.
C1=100uF and no resistance drawn in series with the signal source looks bad. The peak base current could burn the the transistor. Hopefully you have there some resistance in series, say at least 50 Ohm. I guess 1uF capacitor could work well in demonstration purposes and at 11.6kHz its higher reactance gives some protection to the transistor. In addition you can use a ceramic or plastic 1uF capacitor. Electrolytic isn't a good idea without polarizing DC.
You seem to have no load. The LC circuit can develop peak AC voltage = your DC supply if you happen to hit the resonant frequency and there's remarkably low losses in the coil. That AC is summed to the DC when you calculate transistor's peak Vce. Just 9V supply makes total 18V peak to your transistor which isn't fatal but if you have more than 9V beware.