Geo4.118

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Find the equation to the circle which is orthogonal to each of x^{2}+y^{2}+4x+2y+1=0,2x^{2}+2y^{2}+8x+6y-3=0,x^{2}+y^{2}+6x-2y-3=0\,

The radical axis of the first two circles and 2 and 3 are

-y+{\frac  {5}{2}}=0,-2x+5y+{\frac  {3}{2}}=0\,

y={\frac  {5}{2}},2x-5y={\frac  {3}{2}}\,

y={\frac  {5}{2}},x=7\,

The length of the tangent from the radical centre to the first circle is

{\sqrt  {{\frac  {25}{4}}+49+28+5+1}}\,

Now the equation of the circle is

(x-7)^{2}+(y-{\frac  {5}{2}})^{2}={\frac  {25}{4}}+49+28+5+1\,

x^{2}-14x+49+y^{2}-5y+{\frac  {25}{4}}={\frac  {25}{4}}+49+28+5+1\,

x^{2}+y^{2}-14x-5y-34=0\,


Main Page:Geometry:Circles