Given the circle, midpoint $O(0,0)$ and radius $r$; on this circle, the variable point $A(r\cos \alpha,r\sin \alpha)$. Given the fixed point $K(k,0)$; through $K$, the fixed line $y=m(x-k)$, cutting the circle in the points $P(\frac{km^{2}-w}{m^{2}+1},-\frac{m(w+k)}{m^{2}+1})$ and $Q(\frac{km^{2}+w}{m^{2}+1},\frac{m(w-k)}{m^{2}+1})$ with $w=\sqrt{r^{2}(m^{2}+1)-k^{2}m^{2}}$. The $\triangle APQ$ has an orthocenter $H(r\cos \alpha +\frac{2km^{2}}{m^{2}+1},r\sin \alpha -\frac{2km}{m^{2}+1})$. Eliminating the parameter $\alpha$, we find the circle $(x-\frac{2km^{2}}{m^{2}+1})^{2}+(y+\frac{2km}{m^{2}+1})^{2}=r^{2}$.