Find the equations of transverse common tangents for two circles
x2 + y2 + 6x − 2y + 1 =0 , x2 + y2 − 2x − 6y + 9 = 0
35x2 + 12xy − 18x = 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)
c1c2=√16+4=2√5=4.47
r1=√g2+f2−c=√9+1−1=3
r2=√g2+f2−c=√1+9−9=1
c1c2>r1+r2
Both the circle lies outside each other.
Let the intersection point of tarnsverse common tangent be P. P divides c1 & c2 internelly in the ratioo of
r1:r2 here ,in the ratio of 3:1.Co-ordinate of P(x,y)=(m1x2+m2x1m1+m2,m1y2+m2y1m1+m2)
=(3×1+1×1−33+4,3×3+1×13+1)
=(0,52)
Equation of transvers common tangent or pair of tangent from the point P is
T2=SS1 - - - - -(3)
T=xx1+yy1+3(x+x1)−(y+y1)+1
=x × 0y52+3(x+0)−(y+52)+1
=52y+3x−y−52+1=32y+3x−32
32(y+6x−1)
s1=(0)2+(52)2+6(0)−2(52)+1
=254−102+1=25−20+44=94
Substituting values of T & s1 in equation (3)
94(y+6x−1)2=(x2+y2+6x−2y+1)×94
35x2+12xy−18x=0
Equation of transverse common tangents
35x2+12xy−18x=0