Answer
$$(2,2);(2,-2);(-2,2);(-2,-2)$$
Work Step by Step
$$\begin{cases}x^2+y^2=8\\x^2-y^2=0\end{cases}$$
Add both equations:
$$x^2+y^2+x^2-y^2=8+0$$
$$2x^2=8$$
$$x^2=4$$
$$x=\pm\sqrt4=\pm2$$
Find corresponding $y$ values for each $x$ value:
$$(2)^2-y^2=0~~~\to~~~ y^2=4~~~\to~~~y=\pm2$$
$$(-2)^2-y^2=0~~~\to~~~ y^2=4~~~\to~~~y=\pm2$$
Solutions are:
$$(2,2);(2,-2);(-2,2);(-2,-2)$$
