(5) If the line $a x-b y+1=0$ bisects the circumference of the circle $C: x^{2}+y^{2}+2 x-4 y+1=$ 0, then the range of values for $a b$ is ( ).
(A) $\left(-\infty, \frac{1}{4}\right]$
(B) $\left(-\infty, \frac{1}{8}\right]$
(C) $\left(0, \frac{1}{4}\right]$
(D) $\left(0, \frac{1}{8}\right]$
(5) B Hint: Since the circumference of the circle is bisected by the line, the center of the circle lies on the line, and the coordinates of the center are $(-1,2)$, so we have $-a-2 b+1=0$, hence $a+2 b=1$, thus
$$
a b=(1-2 b) b=-2 b^{2}+b=-2\left(b-\frac{1}{4}\right)^{2}+\frac{1}{8} \leqslant \frac{1}{8},
$$

with equality holding if and only if $b=\frac{1}{4}$, therefore $a b \in\left(-\infty, \frac{1}{8}\right]$.