Geo4.53

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Find the locus of the point whose polars with respect to the circles x^{2}+y^{2}-4x-4y-8=0\, and x^{2}+y^{2}-2x+6y-2=0\, are mutually perpendicular.

Let P(x1,y1) be a point

Equation to polar of P w.r.t the first circle is

xx_{1}+yy_{1}-2(x+x_{1})-2(y+y_{1})-8=0\,

x(x_{1}-2)+y(y_{1}-2)-(2x_{1}+2y_{1}+8)=0\,

Equation of polar w.r.t the second circle is

xx_{1}+yy_{1}-(x+x_{1})+3(y+y_{1})-2=0\,

x(x-x_{1})+y(y_{1}+3)-(x_{1}-3y_{1}+2)=0\,

Given these two lines are perpendicular, hence

(x_{1}-2)(x_{1}-1)+(y_{1}-2)(y_{1}+3)=0\,

The locus of P is

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


Main page:Geometry:Circles