Geo4.83

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Find the equations of common tangents to the circles x^{2}+y^{2}-4x-10y+28=0,x^{2}+y^{2}+4x-6y+4=0\,.

Centre and radii of the given two circles are

A(2,5),r_{1}={\sqrt  {4+25-28}}=1,B(-2,3),r_{2}={\sqrt  {4+9-4}}=3\,

AB={\sqrt  {(2+2)^{2}+(5-3)^{2}}}=2{\sqrt  {5}},r_{1}+r_{2}=4\,

AB>(r_{1}+r_{2})\,

Then two circles don't intersect with eachother an done stay outside the other.

Let C2=the external centre of simillitude. C2 divides AB in the ratio 1:-3

C_{2}=\left({\frac  {1\cdot (-2)+(-3)2}{1-3}},{\frac  {1\cdot 3+(-3)5}{1-3}}\right)\,

C_{2}=\left(4,6\right)\,

Equation of the pair of common tangents through C2 to the circle one is

S_{1}=x(4)+y(6)-2(x+4)-5(y+6)+28=2x+y-10\,

S_{{11}}=4^{2}+6^{2}-4(4)-10(6)+28=4\,

[S_{1}]^{2}=SS_{{11}}\,

(2x+y-10)^{2}=(x^{2}+y^{2}-4x-10y+28)4\,

4x^{2}+y^{2}+100+4xy-20y-40x=4x^{2}+4y^{2}-16x-40y+112\,

-3y^{2}+4xy-24x+20y-12=0\,

3-y^{2}+20y-12-24x+4xy=0\,

(y-6)(3y-2)-4x(6-y)=0\,

(y-6)(4x-3y+2)=0\,

y-6=0,4x-3y+2=0\,

Main Page:Geometry:Circles