Answer
$(0,2),(0,-2)$
Work Step by Step
$x^{2}+4y^{2} = 16$ Equation $(1)$
$x^{2}+y^{2} =4$ Equation $(2)$
Subtract Equation $(2)$ from Equation $(1)$
$x^{2}+4y^{2} - (x^{2}+y^{2} )= 16-4$
$x^{2}+4y^{2} - x^{2}-y^{2} =12$
$3y^{2}= 12$
$y^{2}= 4$
$y=±2$
$y = 2$ or $y=-2$
Substitute $y$ values in Equation $(2)$ to get corresponding $x$ values.
Let $y = 2$
$x^{2}+y^{2} =4$
$x^{2}+2^{2} =4$
$x^{2}+4=4$
$x^{2}=0$
$x=0$
Let $y = -2$
$x^{2}+y^{2} =4$
$x^{2}+(-2)^{2} =4$
$x^{2}+4=4$
$x^{2}=0$
$x=0$
$(0,2),(0,-2)$ satisfy given equations. Solution set: $\{(0,2),(0,-2)\}$
