Система уравнений x^2+y^2=16,x-y=4

$\left \{ {{ x^{2} +y^2=16} \atop {x-y=4}} \right.$
$\left \{ {{ (x-y)^2+2xy=16} \atop {x-y=4}} \right.$
$\left \{ {{ 4^2+2xy=16} \atop {x-y=4}} \right.$
$\left \{ {{16+2xy=16} \atop {x-y=4}} \right.$
$\left \{ {{2xy=0} \atop {x-y=4}} \right.$
$\left \{ {{xy=0} \atop {x=4+y}} \right.$
$\left \{ {{y(4+y)=0} \atop {x=4+y}} \right.$
$y=0$ или $y+4=0$
$y=-4$
$\left \{ {{y=0} \atop {x=4}} \right.$ или $\left \{ {{y=-4} \atop {x=0}} \right.$

Ответ: (0;-4); (4;0)