Conditions for maximum power transfer in an AC circuit
Maybe I'm a bit late to the party, but this question does not have a correct answer yet.

The trick is to first calculate a reactance to be put in parallel with the load, such that the resulting impedance has a real part equal to 150 ohm. This happens when a parallel reactance of -2000 ohm is chosen. The imaginary part of the resulting impedance will then be -274 ohm, which can be compensated by a positive reactance of 274 ohm.

A capacitor of 116 pF has a reactance of -2000 ohm at 685 kHz, an inductor of 63.7 uH will provide the reactance of 274 ohm.

The result is the following network:







simulate this circuit &ndash; Schematic created using CircuitLab

Of course the die hard RF guys would have used a Smith chart :)