Find the equations of direct common tangents for two circles
x2 + y2 + 6x − 2y + 1 = 0, x2 + y2 − 2x − 6y + 9 = 0
3y2 − 4xy + 16x − 12y = 0
Given circles,
x2+y2+6x−2y+1=0 - - - - - - (1)
x2+y2−2x−6y+9=0 - - - - - - (2)
Let the circle be c1&c2 and radii r1&r2 of circle (1) and (2) respectively
c1(−3, 1)
c2(1,3)
r1=√g2+f2−c=√9+1−1=3
r2=√g2+f2−c=√1+9−9=1
c1c2=√16+4=2√5=4.47
c1c2>r1+r2
Both the circle lies outside each other.
Let intersection point of direct common tangent be Q. Q divides c1 & c2
externally in the ratio of r1:r2 here, it divides in the ratio of 3:1. Co-ordinates of
Q(x,y)=(m1x2+m2x1m1+m2,m1y2+m2y1m1+m2)
=(3×1+1×1(−3)3−1,3×3−1×(1)3−1)
=(62,82)=(3,4)
Equation of direct common tangent or pair of tangent from the external point
T2=SS1
T=xx1+yy1+3(x+x1)−(y+y1)+1
=3x+4y+3(x+3)−(y+4)+1
=6x+3y+6
Substituting values of T & s1 in equation (3)
(6x+3y+6)2=(x2+y2+6x−2y+1) (32+42+18−8+1)
=(x2+y2+6x−2y+1)(36)
36x2+9y2+36+36xy+36y+72x=36x2+36y2+216x−72y+36
−27y2+36xy−144x+108y=0
3y2−4xy+16x−12y=0
Equation of transverse common tangents
3y2+4xy+16x−12y=0