Geo4.111

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Find the equation of the circle which is orthogonal to x^{2}+y^{2}+2x+8=0,x^{2}+y^{2}-8x+8=0\, and which touches the line x-y+4=0\,

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

Let the given equations be 2,3,4.

Given that 1 cuts 2 and 3,orthogonally,hence

2g(1)+2f(0)=c+8,2g(-4)+2f(0)=c+8\,

2g=c+8,-8g=c+8\,

Solving these two equations,we get

g=0,c=-8\,

Given that 1 touches the line x-y+4=0\,

{\frac  {|-g+f+4|}{{\sqrt  {2}}}}={\sqrt  {g^{2}+f^{2}-c}}\,

Now substituting the values of g and c,we get

(f+4)^{2}=2(f^{2}+8)\,

f^{2}-8f=0,f=0,f=8\,

Therefore the required equations are

x^{2}+y^{2}=8,x^{2}+y^{2}+16y-8=0\,

Main Page:Geometry:Circles