1. Write both curves in standard circle form

The given curves are:

$x^2-6x+y^2+8=0$ $x^2-8y+y^2+16-k=0,\, k>0$

2. First circle

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

Complete the square in $x$:

$x^2-6x=(x-3)^2-9$

So,

$(x-3)^2-9+y^2+8=0$ $(x-3)^2+y^2=1$

Hence, the first circle has:

• centre $C_1=(3,0)$
• radius $r_1=1$

3. Second circle

$x^2-8y+y^2+16-k=0$

Complete the square in $y$:

$y^2-8y=(y-4)^2-16$

Thus,

$x^2+(y-4)^2-16+16-k=0$ $x^2+(y-4)^2=k$

Hence, the second circle has:

• centre $C_2=(0,4)$
• radius $r_2=\sqrt{k}$

4. Condition for two circles to touch

Distance between centres:

$d=\sqrt{(3-0)^2+(0-4)^2}=\sqrt{9+16}=5$

For two circles to touch each other, either:

• externally: $d=r_1+r_2$
• internally: $d=|r_2-r_1|$

Since $r_1=1$ and $r_2=\sqrt{k}$:

External touching

$5=1+\sqrt{k}$ $\sqrt{k}=4$ $k=16$

Internal touching

$5=|\sqrt{k}-1|$

Since $\sqrt{k}>0$, the feasible case is

$\sqrt{k}-1=5$ $\sqrt{k}=6$ $k=36$

(The other possibility $1-\sqrt{k}=5$ is impossible.)

5. Largest value of $k$

The two possible values are:

$k=16,\;36$

So the largest value is

$\boxed{36}$

6. Comparison with stored correct answer

Stored correct answer: $36$

This matches the derived answer.