Given the midpoint $M(0,0)$ of the side $AB$. Given the bases of the altitudes $H_{1}(a,b)$ and $H_{2}(c,d)$, point $H_{1} \in AC$ and point $H_{2} \in BC$. Condition! $AH_{2} \bot BC \Rightarrow \triangle H_{2}AB$ is rectangular and $H_{2}M =\frac{1}{2}AB =r$. $BH_{1} \bot AC \Rightarrow \triangle H_{1}AB$ is rectangular and $H_{1}M =\frac{1}{2}AB =r$. Then $H_{2}M =H_{1}M =r$ and $a^{2}+b^{2}=c^{2}+d^{2}=r^{2}$. The points $H_{1},H_{2},A,B$ are lying on a circle with radius $r$. $A(r\cos \alpha,r\sin \alpha)$ and $B(-r\cos \alpha, -r\sin \alpha)$. The lines $AH_{2}$ and $BH_{1}$ cut in the orthocenter of $\triangle ABC$. Eliminating the parameter $\alpha$, we find the circle $x^{2}+y^{2}-\frac{2(ac-bd+r^{2})}{a+c}\cdot x-\frac{2(ad+bc)}{a+c}\cdot y+r^{2}=0$, the circle going through $H_{1}$ and $H_{2}$.