• 试题 ID 13653


曲线 $\left\{\begin{array}{l}x^2+y^2+z^2=4, \\ x^2+y^2=2 x\end{array}\right.$ 在点 $(1,1, \sqrt{2})$ 处的法平面方程为
A $\sqrt{2} x-y=0$.
B $\sqrt{2} x-z=0$.
C $\sqrt{2} x-y=\sqrt{2}-1$.
D $\sqrt{2} x-z=\sqrt{2}-1$.
E
F
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