Given circles are $S_1\equiv x^2+y^2-14x+6y+33=0$

$S_2\equiv x^2+y^2-a^2=0$

Here, $r_1+r_2<c_1{c}_2$ as the circles have $4$ common tangents.

$c_1=\left(7,-3\right); c_2=\left(0,0\right)$

$c_1{c}_2=\sqrt{{\left(7-0\right)}^2+{\left(-3-0\right)}^2}=\sqrt{7^2+3^2}=\sqrt{49+9}=\sqrt{58}$

$r_1=\sqrt{7^2+{\left(3\right)}^2-33}=\sqrt{49+9-33}=\sqrt{25}=5$

$r_2=\sqrt{a^2}=a$

$r_2+r_2=5+a<\sqrt{58}$

$\Rightarrow {\left(5+a\right)}^2<58$

The above equation satisfies for only two values of $a$ i.e., $1, 2$.