There are two circles whose equations are $x^2+y^2=9$

and $x^2+y^2-8x-6y+n^2=0, n\in Z$  having exactly two common 

tangents. The number of possible values of $n$, is 

The coordinates of the centres and radii of the

circles are:

Centres:  $C_1(0,0)C_2(4,3)$

Radii   $r_1=3r_2=\sqrt{25-n^2},-5<n<5$

Given circles will have exactly two common tangents, if 

$\begin{array}{l} \left|r_1-r_2\right|<C_1{C}_2<r_1+r_2\\ \Rightarrow \left|3-\sqrt{25-n^2}\right|<5<3+\sqrt{25-n^2}\end{array}$

Clearly, $\left|3-\sqrt{25-n^2}\right|<5$  is true for all $n\in (-5,5)$

Now, 

      $\begin{array}{r} 5<3+\sqrt{25-n^2}\\ \Rightarrow 2<\sqrt{25-n^2}\end{array}$

$\begin{array}{l}\Rightarrow 4<25-n^2\\ \Rightarrow n^2-21<0\\ \Rightarrow -\sqrt{21}<n<\sqrt{21}\Rightarrow n=\pm 4,\pm 3,\pm 2,\pm 1,0.\end{array}$

Hence, $n$can take $9$integral values.