Question

A fixed circle is cut by circles passing through two fixed points $A(x_1,y_1)andB(x_2,y_2).$ Prove that the chord of intersection of the fixed circle with any of the circles passes through a fixed point.

Answer 1

Let $s\equiv x^2+y^2+2gx+2fy+c=0$
The equation of any circle passing through two fixed points A and B is $s+\lambda p=0$
i.e. , Circle on $AB$ as diameter $+\lambda \left(lineAB\right)=0$
$(x-x_1)(x-x_2)+(y-y_1)(y-y_2)$
$(x-x_1)(x-x_2)+(y-y_1)(y-y_2)+\lambda \left|\begin{array}{ccc}x& y& 1\\ x_1& y_1& 1\\ x_2& y_2& 1\end{array}\right|=0$ ....(2)
Common chord of circles (1) and (2) is given by $S_1-S_2=0$
or $(x^2+y^2+2gx+2fy+c)-(x^2+y^2)-x(x_1+x_2)-y(y_1+y_2)+x_1{x}_2+y_1{y}_2+\lambda =0$
or $x(x_1+x_2)+2g-y(y_1+y_2)+2f+x_1{x}_2+y_1{y}_2+\lambda =0$
Above is of the form $p-\lambda Q=0$ which represents a family of lines passing through the intersection of two fixed lines which is fixed point.