Answer
$(0,-2),(-2,0)$
Work Step by Step
$x^{2}+y^{2}=4$ Equation $(1)$
$x+y = -2$ Equation $(2)$
From Equation $(2)$
$y = -2-x$ Equation $(3)$
Substitute $y = -2-x$ in Equation $(1)$
$x^{2}+y^{2}=4$
$x^{2}+(-2-x)^{2}=4$
[Using $(a+b)^{2}=a^{2}+2ab+b^{2}$
$(-2)^{2}+2(-2)(-x)+(-x)^{2} = 4+4x+x^{2}$]
$x^{2}+x^{2}+4x+4=4$
$2x^{2}+4x+4-4=0$
$2x^{2}+4x=0$
$2x(x+2)=0$
$x=0$ or $x=-2$
Substitute $x$ values in Equation $(3)$
Let $x=0$
$y = -2-x$
$y = -2-0$
$y = -2$
Let $x=-2$
$y = -2-(-2)$
$y = -2+2$
$y = 0$
$(0,-2),(-2,0)$ both satisfies the equations $(1)$ and $(2)$
So, solutions are $(0,-2),(-2,0)$
