If the circles given by $S\equiv x^2+y^2-14x+6y+33=0$ and $S^{\mathrm{\prime }}\equiv x^2+y^2-a^2=0(a\in N)$

have 4 common tangents, then the possible number of circles $S^{\mathrm{\prime }}=0$ is 

Centres are $C_1=(7,-3),C_2=(0,0)$ ; Radii are $r_1=\sqrt{49+9-33}=5,r_2=a$

Given  circle shave 4 common tangents $\Rightarrow C_1{C}_2>r_1+r_2\Rightarrow \sqrt{49+9}>5+a\Rightarrow a+5<\sqrt{58}$

The positive integral values are 1,2.   $\therefore$ The number of such circles = 2.