The lines $k x+2 y-4=0$ and $5 x-2 y-4=0$ are conjugate with respect to the circle
$x^2+y^2-2 x-2 y+1=0$
Then, necessary condition
$\begin{aligned}
& r^2\left(a_1 a_2+b_1 b_2\right)=\left(a_1 g+b_1 f-c_1\right) \\
& \left(a_2 g+b_2 f-c_2\right)
\end{aligned}$
From the above equation,
$\begin{aligned}
& a_1=k, \quad b_1=2 \quad c_1=-4 \\
& a_2=5, \quad b_2=-2 \quad c_2=-4 \\
& g=-1, \quad f=-1 \quad c=1 \\
& r^2=g^2+f^2-\mathrm{c}=1+1-1=1
\end{aligned}$
Now, by using equation (i)
$\begin{aligned}
& 1(5 k-1)=(-k-2+1)(-5+2+4) \\
& 5 k-4=-k+2 \\
& 6 k=6 \\
& \Rightarrow \quad k=1
\end{aligned}$