I'm not sure which part of the circuit you are asking for help for.
Is it the detection of 10.55V DC or is it how to make the isolated contact?
Or both?
Could you desribe it in more detail.
Which voltage and current should the "dry contact" output switch?
You can use a standard realy, reed relay, opto-coupler, photovoltaic relay or similar as dry contact, depengng of the task.
The more detailed you describe the task of your circuit the better answers you will get.
To detect the 10.55V supply-voltage you could use comparator with the inverting input connected to a standard 6.8V zener from GND and a current feeding resistor connected to the supply-voltage.
At the non-inverting input you should connect two resistors as a voltage divider from the supply voltage to GND.
The two resistor values should be chosen so you have 6.8V at in the middle of the resistors when the supply-voltage is 10.55V.
Then the output of the comparator will switch from GND to VCC when the voltage reaches 10.55V.
The circuit should look something like this, but Vcc and DC Input should both be connected to the source you want to detedt if this is the only supply available:
Description of the circuit:
While the voltage is below 6.8V the zener will not conduct and the inverting input would therefore be pulled to the supply-voltage.
The voltage at the non-inverting input will be lower than at the inverting input because of the voltage-divider. This will keep the comparator output at GND.
When the input-voltage reaches 6.8V the inverting input will stay at this voltage.
If the resistors values in the voltage-divider is chosen correctly the voltage at the non-inverting input will reach 6.8V when the input-value reaches 10.55V. This will make the comparator output go high and you can use this to turn on your "dry contact" relay, opto-cupler or similar.
You can use a potentiometer to adjust the voltage-divider relationship if you want to fine-tune at which input-voltage the comparator output should go high.