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The origin is a limiting point of a system of coaxal circles of which x^{2}+y^{2}+2gx+2fy+c=0\, is a member.Show that the equations of the circles of the orthogonal system are (x^{2}+y^{2})(g+kf)+c(x+ky)=0\, for different values of k.

Given a member of a coaxal system and (0,0) is a limiting point.

Therefore,the limting point circle equation is x^{2}+y^{2}=0\,

The equation to the radical axis is


Any circle of the coaxal system is

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

Let the equation of the circle cutting the system of coaxal circles orthogonally be

x^{2}+y^{2}+2ax+2by+d=0\,. the equation be 2.


2(a)(g\lambda )+2b(f\lambda )=c\lambda +d\,

The circle 2 passes through the limiting points of the coaxal system, then we have d=0.


2ag+2bf=c,c=2a[g+{\frac  {b}{a}}f]\,

Let {\frac  {b}{a}}=k\,

2a(g+kf)=c,2a={\frac  {c}{g+kf}}\,

2b=2a({\frac  {b}{a}})=2ak={\frac  {ck}{g+kf}}\,

Therefore, the euquation to the orthogonal system is

x^{2}+y^{2}+{\frac  {c}{g+kf}}+{\frac  {ck}{g+kf}}=0\,

(g+kf)(x^{2}+y^{2})+c(x+ky)=0\, for different values of k.

Main page:Geometry:Circles