If the curves,

The given equation of circle

$x^{2}-6 x+y^{2}+8=0$

$(x-3)^{2}+y^{2}=1$ .........(1)

So, centre of circle (1) is $C_{1}(3,0)$ and radius $r_{1}=1$.

And the second equation of circle

$x^{2}-8 y+y^{2}+16-k=0(k>0)$

$x^{2}+(y-4)^{2}=(\sqrt{k})^{2}$........(2)

So, centre of circle (2) is $C_{2}(0,4)$ and radius $r_{2}=\sqrt{k}$ Two circles touches each other when

$C_{1} C_{2}=\left|r_{1} \pm r_{2}\right| \quad \Rightarrow \quad 5=|1 \pm \sqrt{k}|$

Distance between $C_{2}(3,0)$ and $\mathrm{C}_{1}(0,4)$ is

either $\sqrt{k}+1$ or $|\sqrt{k}-1| \quad\left(C_{1} C_{2}=5\right)$

$\Rightarrow \quad \sqrt{k}+1=5 \quad$ or $\quad|\sqrt{k}-1|=5$

$\Rightarrow \quad k=16$ or $k=36$

Hence, maximum value of $k$ is 36

The given equation of circles

$x^{2}-6 x+y^{2}+8=0$

$\Rightarrow(x-3)^{2}+y^{2}=1$