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Find the equation of the circle which cuts orthogonally three circles x^{2}+y^{2}-4x-2y+6=0,x^{2}+y^{2}-2x+6y=0\, and x^{2}+y^{2}-12x+2y+30=0\,

Let the equation of the required circle be

x^{2}+y^{2}+2gx+2fy+c=0\,.Let this equation be 1.

The given circles be 2,3,4.

Equation 1 cuts 2,3,4 orthogonally,

2g(-2)+2f(-1)=c+6,4g+2f=-c-6\,. equation 5.

2g(-1)+2f(3)=c+0,-2g+6f=c\,. equation 6

2g(-6)+2f(1)=c+30,-12g+2f=c+30\,. equation 7.

Substituting 6 in 5 and 7

4g+2f=2g-6f-6,2g+8f=-6,g+4f=-3\,.equation 8

-12g+2f=-2g+6f+30,-10g-4f=30,5g+2f=-15\,. equation 9

Solving 8 and 9



The equation of the required circle is



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