If the circles $x^2+y^2-2x-2(3+\sqrt{7})y+8+6\sqrt{7}=0$and $x^2+y^2-8x-6y+k^2=0,k\in Z$, have exactly two common tangents, then the number of possible values of k is
 

$$\begin{gathered} r_{{ }_1}=\sqrt{1^2+{(3+\sqrt{7})}^2-(8+6\sqrt{7})}=3 \\ \begin{array}{l}{r}_{{ }_2}=\sqrt{4^2+3^2-k^2}=\sqrt{25-k^2}\\ c_1{c}_2=\sqrt{{(4-1)}^2+{(3-3-\sqrt{7})}^2}=4\\ c_1{c}_2<r_{{ }_1}{r}_{{ }_2}\\ \Rightarrow 4<3+\sqrt{25-k^2}\Rightarrow 1<\sqrt{25-k^2}\\ \Rightarrow 1<25-k^2\Rightarrow k^2<24\\ \Rightarrow k=0\pm 1,\pm 2,\pm 3,\pm 4[\because k\in z]\end{array} \end{gathered}$$