Drew up some equivalent circuit diagrams to see what it would take to get down to 83 ohms. Assumed the DC case: all inductors/coils were shorts, caps were opens, drew in transistors as essentially switches that are either totally open or totally shorted (I realize this is not realistic, but this is back of the envelope).
The "perfect" circuit, where all transistors are open, gave 8.8k ohm to ground.
When I assumed the oscillator was shorted, the equivalent resistance ended up ~3.5k ohm.
If FET103 was completely shorted source-to-drain, the resistance got down to 213 ohms, and FET101 or FET102 shorted gave approximately 120 ohms.
The only way to get down to the 83 ohm range was to assume two FETs were shorted: that got down to 74 ohm. Could be any combination of the three, and since that got us below the 83 ohm point we can assume the more realistic, physical case where the short isn't complete and instead merely a low resistance, which would of course bump us up a little into the 80 ohm range.
I'm probably overthinking it, I guess, and there are other possibilities (a shorted or even just leaky ceramic cap throws most of my assumptions out the window as it might provide an alternate low- resistance path to ground which could drive down the equivalent resistance quicker).
Would checking the source-drain resistance of the FETs be of value, or would the other circuitry in the tuner make the values so distorted as to be useless?
Or just bite the bullet and order some FETs?