Hệ đã cho tương đương với: $\left\{\begin{array}{l}(x^2-2)^2+(y-3)^2=4\\(x^2-2)(y-3)+4(x^2+y-5)-8=0\end{array}\right.$
Đặt: $\left\{\begin{array}{l}u=x^2-2\\v=y-3\end{array}\right.$, nệ trở thành: $\left\{\begin{array}{l}u^2+v^2=4\\uv+4(u+v)-8=0\end{array}\right.$
Đặt: $\left\{\begin{array}{l}S=u+v\\P=uv\end{array}\right. (S^2\ge4P)$, ta có hệ: $\left\{\begin{array}{l}S^2-2P=4\\P+4S-8=0\end{array}\right.\Leftrightarrow \left\{\begin{array}{l}S=2\\P=0\end{array}\right.$
Từ đó ta có: $\left[\begin{array}{l}\left\{\begin{array}{l}u=0\\v=2\end{array}\right.\\\left\{\begin{array}{l}u=2\\v=0\end{array}\right.\end{array}\right.\Leftrightarrow \left[\begin{array}{l}\left\{\begin{array}{l}x=\pm\sqrt2\\y=5\end{array}\right.\\\left\{\begin{array}{l}x=\pm2\\y=3\end{array}\right.\end{array}\right. $