If S≡x2+y2+2gx+2fy+c=0.
C≡(−g,−f) and r=g2+f2−c.
The given circles are x2+y2−18x−15y+131=0
⇒C1(9,215),r1=81+4225−131=25
and x2+y2−6x−6y−7=0
⇒C2(3,3),r2=9+9+7=5
⇒d=C1C2=(9−3)2+(215−3)2=36+481=215
⇒r1+r2=25+5=215
⇒C1C2=r1+r2
∴ Circles touch each other externally, 3 common tangents.
Hence this is the correct option.