$$
\begin{aligned}
&\text { Given, } x^{2}+y^{2}=16 x \\
&\Rightarrow \quad x^{2}-16 x+y^{2}=0 \\
&\Rightarrow \quad x^{2}-2(x) 8+(8)^{2}-(8)^{2}+y^{2}=0 \\
&\Rightarrow \quad(x-8)^{2}+y^{2}=64
\end{aligned}
$$
$\therefore$ Centre $=(8,0)$ and radius $=8$
Since, the straight line $3 x+4 y=k$ touches the circle $x^{2}+y^{2}=16$, therefore the length of perpendicular from the centre $(8,0)$ to the straight line $3 x+4 y-k=0$ is equal to the radius of the circle.
$$
\begin{aligned}
\text { i.e., } & 8 &=\left|\frac{3(8)+4(0)-k}{\sqrt{9}+16}\right| \\
\Rightarrow & 8 &=\left|\frac{24-k}{\sqrt{25}}\right| \\
\Rightarrow & 8 &=\frac{|24-k|}{5} \\
\Rightarrow & & 24-k \mid &=40
\end{aligned}
$$
Taking positive sign, we get
$$
\begin{aligned}
\Rightarrow & 24-\mathrm{k} &=40 \\
\Rightarrow & &-\mathrm{k} &=40-24=16 \\
\mathrm{k} &=-16
\end{aligned}
$$
Taking negative sign, we get
$$
\begin{aligned}
& \Rightarrow & 24-k &=-40 \\
\Rightarrow & &-k &=-40-24=-64 \\
& \therefore & k &=64 \\
& &=-16,64
\end{aligned}
$$