Geo4.64

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Find the equation of the tangents drawn from the origin to the circle x^{2}+y^{2}+10x+10y+40=0\,

Given circle is S\equiv x^{2}+y^{2}+10x+10y+40=0\,

For the given point (0,0)

S_{1}\equiv 5x+5y+40,S_{{11}}\equiv 40\,

Equation to the pair of tangents from the origin to S=0 is

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

(5x+5y+40)^{2}=(x^{2}+y^{2}+10x+10y+40)40\,

25(x^{2}+y^{2}+2xy+16x+16y+64)=40(x^{2}+y^{2}+10x+10y+40)\,

15x^{2}+15y^{2}-50xy=0\,

3x^{2}-10xy+3y^{2}=0\,

(3x-y)(x-3y)=0\,

Therefore the tangents are

3x-y=0,x-3y=0\,

Main Page:Geometry:Circles