Let $A(0,a),B(-b,0),C(c,0)$, then $H(0,0)$. Choose $O(\lambda,0)$. Points $X(\frac{b(b\lambda-a^{2})}{a^{2}+b^{2}},\frac{ab(b+\lambda)}{a^{2}+b^{2}}\ )$ and $Y(\frac{c(a^{2}+c\lambda)}{a^{2}+c^{2}},\frac{ac(c-\lambda)}{a^{2}+c^{2}}\ )$. Circumcircle $\triangle BHX\ :\ x^{2}+y^{2}+bx-\frac{b\lambda}{a}\cdot y=0$, midpoint $O_{1}(-\frac{b}{2},\frac{b\lambda}{2a})$ and radius $r_{1}=\frac{b}{2a}\sqrt{a^{2}+\lambda^{2}}$. Circumcircle $\triangle CHY\ :\ x^{2}+y^{2}-cx+\frac{c\lambda}{a}\cdot y=0$, midpoint $O_{2}(\frac{c}{2},-\frac{c\lambda}{2a})$ and radius $r_{2}=\frac{c}{2a}\sqrt{a^{2}+\lambda^{2}}$. The line $O_{1}O_{2}\ :\ y=-\frac{\lambda}{a} \cdot x$. On this line, the point $Q(u,-\frac{u\lambda}{a})$. We use $\frac{O_{1}Q}{O_{2}Q}=\frac{r_{1}}{r_{2}}$. Calculating, we find $u=\frac{bc}{b-c}$, independant from the parameter $\lambda$. The locus of the points $Q$ is the line $x=\frac{bc}{b-c}$.