(5) Given the system of equations about $x$ and $y$: $\left\{\begin{array}{l}x^{2}+y^{2}=2 k^{2}, \\ k x-y=2 k\end{array}\right.$ has only one real solution, then the number of real values of $k$ that satisfy this condition is ( ).
(A) 1
(B) 2
(C) 3
(D) 4
(5) $\mathrm{C}$ Hint: If $k=0$, it is obvious that the system of equations has only one solution $(0,0)$, so $k=0$ meets the condition; If $k \neq 0$, then $x^{2}+y^{2}=2 k^{2}$ represents a circle centered at $(0,0)$ with radius $r=\sqrt{2}|k|$, and $k x-y=2 k$ represents a line. According to the given condition, $\frac{|2 k|}{\sqrt{k^{2}+1}}=\sqrt{2}|k|$, which simplifies to $\frac{4 k^{2}}{1+k^{2}}=2 k^{2}$, solving this yields $k= \pm 1$.
In summary, there are 3 real numbers $k$ that meet the condition.