How can I test for a closed circuit without adding excessive resistance?
Don't put your relay in series with the power supply to this input circuit, but put the relay and the input circuit in parallel (on the power supply), so when the 12V DC power comes on for the input circuit, it also energizes the relay for you. 

This way there is very little voltage drop caused by the relay, assuming the power supply has enough power to supply both the relay and the input circuit.

Unless your 12V is always on, and you want some way to test when current is flowing in the main input device, and when you power on the main input device, you want the relay to switch too?  Then you need a current sense on the 12V DC line to switch a relay for you.  In the circuit below, the transistor Q1 turns on whenever there is current flowing in the sense resistor R1.  The specific value of Q1 depends on the current drawn by the relay coil. It would need to be large enough to support the relay current.  







simulate this circuit &ndash; Schematic created using CircuitLab

This is a very simplified circuit too.  If your input device draws a lot of current, it will cause a larger voltage drop across the 10 ohm sense resister, so you will want to make that smaller.  If your input device draws less current you will want to make that larger.   As it is, the 10 ohm sense resistor R1 will cause the transistor to turn on when about 60mA is drawn from the input device.  If your device draws more than 60mA, it will still turn on the relay, but the excess current draw will cause more voltage drop across the 10 ohm R1, so you want to make that smaller. 

For example, if your input device draws 200mA, then 200mA x 10 ohms will causes a 2V drop, so your 12V - 2V is 10V left for the input device, and this may be too much voltage drop for your device. So you will want to reduce the sense resistor to about 3 ohms, so that it only drops 0.6 Volts at 200 mA current. 

Use Ohms law to figure out the R1 resistor value. If you know your device draws 200mA current, then R1 = 0.6 / 0.200 = 3 ohms.  For better turn on, you probably want the relay to turn on at 80% of your device's current draw, so if your device draws 200mA, then 80% of that is 160mA, so choose R1 to turn on the transistor at 160mA instead ==>  R1 = 0.6V / .160 = 3.75 ohms