Geo4.79

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Find the equations of the direct common tangents to the circles x^{2}+y^{2}=1,(x-1)^{2}+(y-3)^{2}=4\,

The centre and radius for the two circles are

A=(0,0),r_{1}=1,B=(1,3),r_{2}=2\,

AB={\sqrt  {(0-1)^{2}+(0-3)^{2}}}={\sqrt  {10}},r_{1}+r_{2}=3\,

From the both,

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

Hence the two circles don't intersect eachother, and each will be outside the other.

Let C2=the external centre

Therefore C2 divides AB in 2:-1

Hence

c_{2}=\left({\frac  {2\cdot 0-1\cdot 1}{2-1}},{\frac  {2\cdot 0-1\cdot 3}{2-1}}\right)\,

C_{2}=(-1,-3)\,

Equations of the direct common tangents through C2 to the circle x^{2}+y^{2}=1\, is

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

(-x-3y-1)^{2}=(x^{2}+y^{2}-1)(1=9-1)\,

x^{2}+y^{2}+1+6xy+2x+6y=9x^{2}+9y^{2}-9\,

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

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

Hence the equations are

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


Main Page:Geometry:Circles