Answer
$\{(-3+\sqrt 7\ i,3+\sqrt 7\ i),(-3-\sqrt 7\ i,3-\sqrt 7\ i) \}$
Work Step by Step
1. From the second equation, we have $y=x+6$
2. Substitute the above in the first equation to get $x^2+(x+6)^2=4$ or $2x^2+12x+36-4=0$ or $x^2+6x+16=0$
3. Use formula, we have $x=\frac{-6\pm\sqrt {6^2-4(16)}}{2}=-3\pm\sqrt 7\ i$
4. Back substitute to get $y=3\pm\sqrt 7\ i$
5. Thus the solution set is $\{(-3+\sqrt 7\ i,3+\sqrt 7\ i),(-3-\sqrt 7\ i,3-\sqrt 7\ i) \}$
