Question
Prove that the locus of the intersection of two equal circles which are described on two sides $EF$ and $EG$ of a triangle as chords is a rectangular hyperbola whose center is the midpoint of $FG$ and which passes through $E$,$F$ and $G$.
Attempt
Consider a triangle $EFG$ with $E(x_1,y_1)$,$F(x_2,y_2)$,$G(x_3,y_3)$ as vertices. Consider two equal circles, $$S_1:=x^2+y^2+2g_1x+2f_1y+c_1=0$$ $$S_2:=x^2+y^2+2g_2x+2f_2y+c_2=0$$ Therefore, the condition of equal radii $(r_1=r_2)$ gives, $$\sqrt {g_1^2+f_1^2-c_1^2}=\sqrt {g_2^2+f_2^2-c_2^2}$$ As $EF$ is the chord of $S_1$, $$x_1^2+y_1^2+2g_1x_1+2f_1y_1+c_1=0$$ $$x_2^2+y_2^2+2g_1x_2+2f_1y_2+c_1=0$$ As $EG$ is the chord of $S_2$, $$x_1^2+y_1^2+2g_2x_1+2f_2y_1+c_2=0$$ $$x_3^2+y_3^2+2g_2x_3+2f_2y_3+c_2=0$$ Let $S_1$ and $S_2$ meet in $(h,k)$. Then we have, $$h^2+k^2+2g_1h+2f_1k+c_1=0$$ $$h^2+k^2+2g_2h+2f_2k+c_2=0$$ What now I have to do is prove that $(h-\frac {x_2+x_3}{2})^2 - (k-\frac {y_2+y_3}{2})^2=A^2$ for some constant $A$. One value of $(h,k)$ is clearly $(x_1,y_1)$. Yet neither I can derive the other value of $(h,k)$ nor prove the desired result, for it looks tremendously tedious. Can anyone help me to make the way out?
Edit
Playing with the slope of the common chord of the circles, I find another linear equation in $h$ and $k$, $$h(f_1-f_2)+k(g_1-g_2)=x_1(f_1-f_2)+y_1(g_1-g_2)$$ As the resulting hyperbola passes through $E$,$F$ and $G$, I have another set of results, $$(x_1-\frac {x_2+x_3}{2})^2 - (y_1-\frac {y_2+y_3}{2})^2=A^2$$ $$(x_2-\frac {x_2+x_3}{2})^2 - (y_2-\frac {y_2+y_3}{2})^2=A^2$$ $$(x_3-\frac {x_2+x_3}{2})^2 - (y_3-\frac {y_2+y_3}{2})^2=A^2$$ Moreover , as suggested in the comment, I would add that this is the case of a conic through 5 points, $E,F,G,$ the reflections of $E$ in the circumcentre of $∆EFG$ and in the midpoint of $FG$.