Geo4.119

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Find the equation to the circle passing through (1,-1) and belonging to the coaxal system determined by the circles x^{2}+y^{2}-6x-6y+4=0,x^{2}+y^{2}-x+2y-3=0\,

The radical axis from the given circles is

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

-5x-8y+7=0\,

5x+8y-7=0\,

Therefore, any circel of the coaxal system is

x^{2}+y^{2}-x+2y-3+\lambda (5x+8y-7)=0\,

If this passes through (1,-1),we get

-4-10\lambda =0,\lambda ={\frac  {-2}{5}}\,

Now the required circle is

x^{2}+y^{2}-x+2y-3-{\frac  {2}{5}}(5x+8y-7)=0\,

Simplifying

5(x^{2}+y^{2})-15x-16y-1=0\,


Main Page:Geometry:Circles