Answer
The solution is at (-2, 0) and (0, -2).
See Graph I for the graphing of the solution.
Work Step by Step
$x^2 + y^2 = 4 \longrightarrow (i)$
$x + y = -2 \longrightarrow (ii)$
From $(ii)$, we have
$x + y = -2$
$x = -y -2 \longrightarrow (iii)$
Sub. $(iii)$ into $(i)$, we have
$(-y-2)^2 + y^2 = 4$
$y^2 + 4y + 4 + y^2 = 4$
$2y^2 + 4y = 0$
$2y(y + 2) = 0$
$y = 0$ or $y = -2$
From $(ii)$,
when $y = 0$, $x = -2$;
when $y = -2$, $x = 0$
The 2 intersection points are at (-2, 0) and (0, -2).
See Graph I for the graphing of the solution.