We have, ‌S1:x2+y2−4x+6y+4=0 ‌C1:(2−3),r1=3 ‌S2:x2+y2+2x−2y−2=0 ‌C2:(−1,1),r2=2 Here, C1C2=r1+r2 So, the point P divides C1C2 in the ratio of 3:2
P(‌
1
5
,‌
−3
5
) Slope of line passing through C1 and C2 is ‌
1+3
−1−2
=‌
4
−3
⇒P:(‌
1
5
,−‌
3
5
) and here only one transverse common tangent which is perpendicular to C1C2 ‌ ⇒ slope of tangent ‌=−‌
