Geo4.120

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Find the equation of the circle belonging to the coaxal system determined by the circles x^{2}+y^{2}-4x-6y-12=0,x^{2}+y^{2}+6x+4y-12=0\, and cuts the circle x^{2}+y^{2}-2x+3=0\, orthogonally.

Radical axis of the first two equations is

(x^{2}+y^{2}-4x-6y-12)-(x^{2}+y^{2}+6x+4y-12)=0\,

-10x-10y=0\,

x+y=0\,

Now the equation of the circle of a coxal system is

x^{2}+y^{2}-4x-6y-12+\lambda (x+y)=0\,

x^{2}+y^{2}-2x(1-{\frac  {\lambda }{2}})-2y(3-\lambda )-12=0\,

This circle cuts the circle x^{2}+y^{2}-2x+3=0\, orthogonally.

Hence,

-(2-\lambda )(-1)+2(-3+\lambda )(0)=-12+3\,

\lambda =-7\,

Therefore the required equation is

x^{2}+y^{2}+5x+8y-12=0\,

Main Page:Geometry:Circles