Question

Решите уравнение.$x^{2}+y^{2}+4x-8y+20=0$

Answer 1

${x}^{2} + {y}^{2} + 4x - 8y + 20 = 0 \ ({x}^{2} + 4x + 4) + ( {y}^{2} - 8y + 16) = 0 \ {(x + 2)}^{2} + {(y - 4)}^{2} = 0 \ x + 2 = 0 \ x = - 2 \ y - 4 = 0 \ y = 4 \ ( - 2;4)$