How to switch separate circuit from doorbell?
What you want is called a "relay".  Ideally you want one that runs from 10 VAC.  That's not a common coil voltage, but 12 V and 6 V are.  You could use a 6 VAC relay with a resistor in series, or full wave rectify the 10 VAC to make close enough to 12 VDC on the capacitor after the diodes.

Either way, the relay is basically a electrically operated mechanical switch.  At minimum, you will probably find a SPDT (single pole double throw) relay.  That gives you the option of using it as a switch that is normally open except when the button is pressed, or normally closed except when the button is pressed.

Here is a schematic of what I'm talking about:



Note the capacitor after the full wave bridge.  This is necessary to keep the average rectified voltage high enough to reliably activate the relay.  The particular relay I'm showing in this example draws 27 mA at 12 V.  The peaks of the 10 VAC waveform will be 14.1 V.  The full wave bridge subtracts two diode drops for about 1.4 V, leaving 12.7 V peaks.  That will be what the cap gets charged up to twice per power line cycle.  For a 12 V average, the voltage on the cap can drop to 11.3 V before the next peak, for a total drop of 1.4 V.  You didn't say what your power line frequency is, so I assumed 50 Hz.  That means the cap will get recharged every 10 ms.  This gives us enough to compute the target cap value:

&nbsp;&nbsp; C = (27 mA)(10 ms)/(1.4 V) = 193 &micro;F

So the common value of 200 &micro;F will work fine.