‌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 ⇒‌=‌
