Có : $x+2y=2$
$\to x = 2-2y$
Khi đó : $x^2+y^2=1$
$\to (2-2y)^2+y^2=1$
$\to 4+4y^2-8y+y^2=1$
$\to 5y^2-8y+3=0$
$\to (y-1).(5y-3)=0$
$\to$ \(\left[{}\begin{matrix}y=1\\y=\dfrac{3}{5}\end{matrix}\right.\)
Khi \(y=1\Rightarrow x=0\)
Khi \(y=\dfrac{3}{5}\Rightarrow x=\dfrac{4}{5}\)