Answer
$(-2,-2)$ and $(2,-2)$
Work Step by Step
$\left\{\begin{aligned}x^{2}+y^{2}&=8\\x^{2}+y^{2}+4y&=0\end{aligned}\right.$
Graphed with desmos.com/calculator.
Substitute 8 for $x^{2}+y^{2}$ in the second equation:
$8+4y=0$
$4y=-8$
$y=-2$
Back-substituting,
$x^{2}+(-2)^{2}=8$
$x^{2}=4$
$x=\pm 2$
Intersections: $(-2,-2)$ and $(2,-2)$
