Question

х^2-8x+y^2+4y+20=0
решите уравнение

Answer 1

$x^2-8x+y^2+4y+20=0 \\ x^2-8x+20+4y+y^2=0 \\ (x-4)^2+4+4y+y^2=0 \\ (x-4)^2+(y+2)^2=0 \\ \left \{ {{x-4 =0} \atop {y+2=0}} \right. \to \left \{ {{x=4} \atop {y=-2}} \right.$