Answer
Graphing: $(-3.464,2),(0,-4),$ and $(3.464,2)$
Algebraically: $(0,-4),(2\sqrt{3},2),(-2\sqrt{3},2)$
Work Step by Step
Graphed with desmos.com/calculator.
$\left\{\begin{array}{ll}
x^{2}+y^{2}=16 & \\
x^{2}-2y=8 & x^{2}=2y+8
\end{array}\right.$
Substitute $x^{2}$ in the first equation
$\begin{aligned}2y+8+y^{2}&=16\\y^{2}+2y-8&=0\end{aligned}$
... two factors of $-8$ whose sum is $+2$ ... are $-2$ and $+4$.
$ (y+4)(y-2) =0$
Back-substituting
$\left[\begin{array}{lll}
y=-4 & ... & x=2\\
x^{2}=2(-4)+8 & & x^{2}=2(2)+8\\
x=0 & & x=\pm\sqrt{12}\\
& & \\
(-4,0) & & (2,\pm 2\sqrt{3})
\end{array}\right]$
Graphing: $(-3.464,2),(0,-4),$ and $(3.464,2)$
Algebraically: $(0,-4),(2\sqrt{3},2),(-2\sqrt{3},2)$
