TS EAMCET 04 May 2019 Shift 1 Solved Paper

Consider the equation
x2+y2−2x+4y−4=0
And
x2+y2+4x−4y+α=0
This implies,
c1=(1,−2)
r1=√1+4+4=3
And,
c2=(−2,2)
r1=√4+4−α=√8−α
Circles have 4 common tangents so,
c1c2&>r1+r2
√(1+2)2+(−2−2)2&>3+√8−α
2&>√8−α
8−α<4
This implies, α>4
Therefore, the least integral value of α is 5