8. Given the equation of circle $\odot C$ is $x^{2}+y^{2}-8 x+15=0$. If there exists at least one point on the line $y=k x-2(k \in \mathbf{R})$ such that the circle with this point as its center and 1 as its radius has a common point with $\odot C$, then the maximum value of $k$ is $\qquad$.
8. $\frac{4}{3}$.

From the problem, we know the center of the circle $C(4,0)$, with a radius of 1.
If the line $y=k x-2$ has at least one point $A\left(x_{0}, k x_{0}-2\right)$, such that the circle $\odot A$ with radius 1 intersects with $\odot C$, i.e., $|A C| \leqslant 1+1=2$, then
$$
|A C|_{\min } \leqslant 2 \text{. }
$$

Thus, the distance from point $C$ to the line $y=k x-2$ is
$$
d=\frac{|4 k-2|}{\sqrt{k^{2}+1}} \leqslant 2 \Rightarrow 0 \leqslant k \leqslant \frac{4}{3} \text{. }
$$

Therefore, the maximum value of $k$ is $\frac{4}{3}$.