Circle

‌c1:x2+y2−2x+4y+1=0
‌⇒(x−1)2+(y+2)2=4
Centre (c1)=(1,−2),r1=2
and c2:x2+y2−4x−2y+4=0
⇒(x−2)2+(y−1)2=1
Centre (c2)=(2,1),r2=1
c1c2=d=√(2−1)2+(1+22.=√10
Let the slope of direct common tangent be m the general line is
y=mx+c
this line has perpendicular distance r1 from c1 and r2 from c2
‌∴|‌

m(1)−(−2)+c

√1+m2

|=2
‌⇒|m+2+c|=2√1+m2
‌‌ and ‌|‌

m(2)−1+c

√1+m2

|=1
‌⇒|2m+c−1|=√1+m2
Put the value of ' c ' from Eq. (i) to Eq. (ii)
2m−1+(−m−2+2√m2+1)=√m2+1
⇒m=3−√m2+1
Squaring both sides, we get
‌m2=‌m2+10
⇒√m2+1‌−6√m2+1
⇒‌m2+1
⇒‌=‌

25

9

⇒‌m2=16∕9
⇒‌m=4∕3