Cho (S1): (x-1)^2+y^2+(z-2)^2=1; (S2): (x+1)^2, A. (S1), (S2) cắt nhau

Câu hỏi :

Cho $\left(S_1\right):\hspace{0.17em}{\left(x-1\right)}^2+y^2+{\left(z-2\right)}^2=1$; $\left(S_2\right):\hspace{0.17em}{\left(x+1\right)}^2+{\left(y+1\right)}^2+z^2=16$. Xác định vị trí tương đối giữa $\left(S_1\right),\left(S_2\right)$.