Given $A(0,0),B(b,0),C(c,a)$. Following the solution in 2007. A line $l$ through $A$ with equation $y=mx$. A circle through $A$ and $B$ with equation $(x-\frac{b}{2})^{2}+(y-\omega)^{2}=\frac{b^{2}}{4}+\omega^{2}$. The line $l_{1}$ cuts this circle in $P_{1}(\frac{b+2m\omega}{1+m^{2}},\frac{m(b+2m\omega)}{1+m^{2}})$. A line $l_{2}$ through $P_{1}$ with slope $\frac{m-\sqrt{3}}{\sqrt{3}m+1}$ ($\angle (l_{1},l_{2})=60^{\circ}$): point $B \in l_{2}$, so $\omega=-\frac{b}{2\sqrt{3}}$, we find the midpoint $D$, see drawing. A circle through $B$ and $C$ with equation $(x-\theta)^{2}+[y-(\frac{a}{2}+\frac{(b-c)\theta}{a}-\frac{b^{2}-c^{2}}{2a})]^{2}=(b-\theta)^{2}+(\frac{a}{2}+\frac{(b-c)\theta}{a}-\frac{b^{2}-c^{2}}{2a})^{2}$. The line $l_{2}$ cuts this circle in the point $P_{2}$. A line $l_{3}$ through $P_{2}$ with slope $\frac{-m-\sqrt{3}}{\sqrt{3}m-1}$ ($\angle (l_{2},l_{3})=60^{\circ}$): point $C \in l_{3}$, so $\theta=\frac{\sqrt{3}[a+\sqrt{3}(b+c)]}{6}$, we find the midpoint $F$, see drawing. Line $l_{3}$ cuts $l_{1}$ in the point $P_{3}$. The center $M(x,y)$ of $\triangle P_{1}P_{2}P_{3}$ $x=\frac{a(\sqrt{3}m^{2}+2m-\sqrt{3})+b(m^{2}-2\sqrt{3}m+3)+c(m^{2}+2\sqrt{3}m+3)}{6(m^{2}+1)}$ and $y=\frac{a(\sqrt{3}m-1)^{2}-b(\sqrt{3}m^{2}-2m-\sqrt{3})+c(m+\sqrt{3})(\sqrt{3}m-1)}{6(m^{2}+1)}$. Eliminating the parameter $m$ was a big problem, but finally we have the circle (in red) with equation $3\sqrt{3} \cdot x^{2}+3\sqrt{3} \cdot y^{2}-2\sqrt{3} \cdot (b+c)x-2\sqrt{3} \cdot ay+ab+\sqrt{3} \cdot bc=0$.

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