The number of common tangents, to the circles x 2 + y 2 - 18 x - 15 y + 131 = 0 and x 2 + y 2 - 6 x - 6 y -…

If $S\equiv x^2+y^2+2gx+2fy+c=0$.

$C\equiv \left(-g,-f\right)$ and $r=\sqrt{g^2+f^2-c}$.

The given circles are $x^2+y^2-18x-15y+131=0$

$\Rightarrow C_1\left(9, \frac{15}{2}\right), r_1=\sqrt{81+\frac{225}{4}-131}=\frac{5}{2}$

and $x^2+y^2-6x-6y-7=0$

$\Rightarrow C_2\left(3, 3\right), r_2=\sqrt{9+9+7}=5$

$\Rightarrow d=C_1{C}_2=\sqrt{{\left(9-3\right)}^2+{\left(\frac{15}{2}-3\right)}^2}=\sqrt{36+\frac{81}{4}}=\frac{15}{2}$

$\Rightarrow r_1+r_2=\frac{5}{2}+5=\frac{15}{2}$

$\Rightarrow C_1{C}_2=r_1+r_2$

$\therefore$ Circles touch each other externally, $3$ common tangents.

Hence this is the correct option.