14.39*. Find the equation of the circumcircle of triangle $A_{1} A_{2} A_{3}$ in barycentric coordinates.
14.39. Let $X$ be an arbitrary point, $O$ the center of the circumcircle of a given triangle, $\boldsymbol{e}_{i}=\overrightarrow{O A_{i}}$, and $\boldsymbol{a}=\overrightarrow{X O}$. If the point $X$ has barycentric coordinates $\left(x_{1}: x_{2}: x_{3}\right)$, then $\sum x_{i}\left(\boldsymbol{a}+\boldsymbol{e}_{i}\right)=\sum x_{i} \overrightarrow{X A_{i}}=\mathbf{0}$, since $X$ is the center of mass of the points $A_{1}, A_{2}, A_{3}$ with masses $x_{1}, x_{2}, x_{3}$. Therefore, $\left(\sum x_{i}\right) \boldsymbol{a}=-\sum x_{i} \boldsymbol{e}_{i}$. The point $X$ lies on the circumcircle of the triangle if and only if $|\boldsymbol{a}|=X O=R$, where $R$ is the radius of this circle. Thus, the circumcircle of the triangle is given in barycentric coordinates by the equation $R^{2}\left(\sum x_{i}\right)^{2}=\left(\sum x_{i} e_{i}\right)^{2}$, i.e., $R^{2} \sum x_{i}^{2}+2 R^{2} \sum_{i<j} x_{i} x_{j}=R^{2} \sum x_{i}^{2}+2 \sum_{i<j} x_{i} x_{j}\left(\boldsymbol{e}_{i}, \boldsymbol{e}_{j}\right)$, since $\left|\boldsymbol{e}_{i}\right|=R$. This equation can be rewritten as $\sum_{i<j} x_{i} x_{j}\left(R^{2}-\left(\boldsymbol{e}_{i}, \boldsymbol{e}_{j}\right)\right)=0$. Note now that $2\left(R^{2}-\left(\boldsymbol{e}_{i}, \boldsymbol{e}_{j}\right)\right)=a_{i j}^{2}$, where $a_{i j}$ is the length of the side $A_{i} A_{j}$. Indeed, $a_{i j}^{2}=\left|\boldsymbol{e}_{i}-\boldsymbol{e}_{j}\right|^{2}=\left|\boldsymbol{e}_{i}\right|^{2}+\left|\boldsymbol{e}_{j}\right|^{2}-2\left(\boldsymbol{e}_{i}, \boldsymbol{e}_{j}\right)=2\left(\boldsymbol{R}^{2}-\left(\boldsymbol{e}_{i}, \boldsymbol{e}_{j}\right)\right)$. In the end, we obtain that the circumcircle of the triangle $A_{1} A_{2} A_{3}$ is given in barycentric coordinates by the equation $\sum_{i<j} x_{i} x_{j} a_{i j}^{2}=0$, where $a_{i j}$ is the length of the side $A_{i} A_{j}$.