Answer
$(-2,0)$
Work Step by Step
$\left\{\begin{aligned}x^{2}+y^{2}&=4\\x^{2}+2x+y^{2}&=0\end{aligned}\right.$
Graphed with desmos.com/calculator.
Rewrite the LHS of the second equation as
$x^{2}+2x+y^{2}=0$
$2x+(x^{2}+y^{2})=0$
... substituting $x^{2}+y^{2}$ with $4...$
$2x+4=0$
$2x=-4$
$x=-2$
Back-substituting:
$(-2)^{2}+y^{2}=4$
$y^{2}=0$
$y=0$
Intersection point is $(-2,0)$
