Given $\triangle ABC\ :\ A(0,a),B(-b,0),C(c,0)$. Midpoint of the circumscribed circle $O(\frac{c-b}{2},\frac{a^{2}-bc}{2a})$. Choose the point $P(p,-q)$, then $R(p,\frac{a^{2}-bc}{a}+q)$. The coordinates of $P$ satisfy the equation of the circle $x^{2}-(c-b)x+y^{2}-\frac{a^{2}-bc}{a} \cdot y-bc=0$. The line $PQ\ :\ y+q=\frac{c}{a}(x-p)$ intersects the circle in the point $Q(-\frac{a^{2}(b-2c+p)-2acq+c^{2}(b-p)}{a^{2}+c^{2}},-\frac{a^{3}q+a^{2}c[b-2(c-p)]-ac^{2}q+bc^{3}}{a(a^{2}+c^{2})}\ )$. The line $RQ\ :\ y-(\frac{a^{2}-bc}{a}+q)=-\frac{b}{a}(x-p)$ intersects the circle in the point $Q(\frac{a^{2}(c-p)+2abq+b^{2}(p-c)}{a^{2}+b^{2}},\frac{a^{3}+a^{2}q+ab[b-2(c-p)]-b^{2}q}{a^{2}+b^{2}})$. Both points $Q$ are the same, so we find $p=\frac{-a^{2}+a(c-b)+bc}{2a}$ and $q=-\frac{a^{2}-a(b+c)-bc}{2a}$. The slope of the line $OP\ :\ m_{OP}=\frac{a(b+c)}{a^{2}-bc}$, the slope of the line $OC\ :\ m_{OC}=\frac{bc-a^{2}}{a(b+c)}$.