Consider the given equation of circles. S1:x2+y2+4x−6y−12=0 Its centre is (-2,3) and radius is r1=5 And, S2:x2+y2−8x+10y+5=0 Its centre is (4,-5) and radius is r2=6 So r1+r2=5+6 =11 And C1C2=√(4+2)2+(−5−3)2 =√36+64 =√100 =10 since r1+r2>C1C2 Therefore, there are only two common tangents to circle S1 and S2