Question

x^2+y^2-2radical5 y + 4 radical5 x +25 = 0

Answer 1

${x}^{2} + {y}^{2} - 2 \sqrt{5} y + 4 \sqrt{5} x + 25 = 0$

${x}^{2} + 4 \sqrt{5}x + \sqrt{5}^{2} \times {2}^{2} + {y}^{2} - 2 \sqrt{5} y + \sqrt{5} ^{2} = 0$

${(x + 2 \sqrt{5} })^{2} + (y - \sqrt{5} ) ^{2} = 0$

$= > x + 2 \sqrt{5} = 0 = > x = - 2√5$

$= > y - \sqrt{5} = 0 = > y = \sqrt{5}$