## College Algebra (10th Edition)

$(-2,-2)$ and $(2,-2)$
\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 $(2,-2)$