AP EAMCET Engineering 20 Apr 2019 Shift 2 Paper

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