Geo4.103

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Find the equation passing through the origin and cutting the circles x^{2}+y^{2}-4x-6y-3=0,x^{2}+y^{2}-16x-6y+4=0\, orthogonally.

Let the equation be

x^{2}+y^{2}+2gx+2fy+c=0\,

This circle passes through the origin,hence

c=0\,

The condition for the required circle and the first circle to cut orthogonally is

From the first circle, g_{1}=-2,f_{1}=-3,c_{1}=-3\,

2(g)(-2)+2(f)(-3)=c-3\,

-4g-6f=-3\,

The condition for the required circle and the second circle to cut orthogonally is

From the second circle, g_{2}=-8,f_{2}=-3,c_{2}=4\,

2(g)(-8)+2(f)(-3)=c+4\,

-16g-6f=4\,

Solving the two equations,we get

-18f=-16,f={\frac  {8}{9}}\,

-16g=20,g={\frac  {-7}{12}}\,

Hence the required circle is

18x^{2}+18y^{2}-21x+32y=0\,

Main Page:Geometry:Circles