How to find the voltage at the circled node
There is 2A flowing counter-clockwise in your circuit, due to the constant current source. Starting at the circle there is a 4V voltage drop through the 2 \$\Omega\$ resistor. The more positive side is at the circle. Continuing counter-clockwise, the battery also has 4V across it. Again, the more positive side is the top, and the more negative on the bottom side of the battery. Since both the resistor and battery are more positive in the same direction, their voltages sum. So the circle is 8V higher than the bottom side of the battery.