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Find the equation of the circle which cuts orthogonally three circles x^{2}+y^{2}+2x+4y+1=0,2x^{2}+2y^{2}+6x+8y-3=0\, and x^{2}+y^{2}-2x+6y-3=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(1)+2f(2)=c+1,2g+4f=c+1\,. equation 5.

2g({\frac  {3}{2}})+2f(2)=c-{\frac  {3}{2}},3g+4f=c-{\frac  {3}{2}}\,. equation 6

2g(-1)+2f(3)=c-3,-2g+6f=c-3\,. equation 7.

(5)-(6) gives

-g={\frac  {5}{2}},g=-{\frac  {5}{2}}\,

(6)-(7) gives 5g-2f={\frac  {3}{2}},f=-7\,

Substituting these values in 5,we get


Therefore the equation of the required circle is


Main Page:Geometry:Circles