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