Answer
See graphs, $\left(0,-4\right),\left(-2\sqrt{3},2\right),\left(2\sqrt{3},2\right)$
Work Step by Step
1. See graphs
2. Take the difference of the two equations $x^2+y^2=16$ and $x^2-2y=8$, we can get $y^2+2y=8$ or $y^2+2y-8=0$
3. Factor to get $(y+4)(y-2)=0$, thus $y=-4, 2$
4. Use the 2nd equation to get $x^2=0, 12$, thus $x=0$ for $y=-4$ and $x=\pm2\sqrt 3$ for $y=2$
5. We can find the intersect(s): $\left(0,-4\right),\left(-2\sqrt{3},2\right),\left(2\sqrt{3},2\right)$
