Решить систему lg(x²+y²)=2 log ₂ x-4=log ₂ 3 - log ₂ y

Решите задачу:

$\left \{ {{lg(x^2+y^2)=2} \atop {log_2(x-4)=log_23-log_2y} \right. \\\\ \left \{ {{lg(x^2+y^2)=lg10^2} \atop {log_2(x-4)=log_2( \frac{3}{y} )}} \right. \\\\ \left \{ {{x^2+y^2=100} \atop {x-4= \frac{3}{y} }} \right.\\\\ \left \{ {{( \frac{3}{y}+4)+y^2=100} \atop {x= \frac{3}{y}+4 }} \right. \\\\ \left \{ {{3+4y+y^3-100=0} \atop {x=\frac{3}{y}+4}} \right. \\ \left \{ {{y^3+4y-77=0} \atop {x= \frac{3}{y}+4}} \right. \\ \left \{ {{y_1=-11;y_2=7} \atop {x_1= \frac{41}{11};x_2= \frac{31}{7} }} \right.$