Geo4.81

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Find the equations to the transverse common tangents of the circles x^{2}+y^{2}+4x+2y-4=0,x^{2}+y^{2}-4x-2y+4=0\,.

The centre and radii of the given circles are

A(-2,-1),r_{1}=3,B(2,1),r_{2}=1\,

AB=2{\sqrt  {5}},r_{1}+r_{2}=4,AB>(r_{1}+r_{2})\,

Hence circles donot intersect with eachother and each will be outside the other,transverse common tangents can be drawn. Let C1 =internal centre of similitude

C1 divides AB in the ratio 3:1

Therefore

C_{1}=\left({\frac  {3\cdot 3+1\cdot (-2)}{3+1}},{\frac  {3\cdot 3+1\cdot (-1)}{3+1}}\right)\,

C_{1}=\left(1,{\frac  {1}{2}}\right)\,

Equation to the pair of transverse common tangents through C1 to circle 1 is

[{\frac  {6x+3y-3}{2}}]^{2}={\frac  {9}{4}}(x^{2}+y^{2}+4x+2y-4)\,

Simplifying

3x^{2}+4xy-8x-4y+5=0\,

(x-1)(3x+4y-5)=0\,

Therefore the equation to the tangents are

x-1=0,3x+4y-5=0\,

Main Page:Geometry:Circles