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Problem 3

Prove that the equation of the circumcircle of the triangle formed by the lines $$ \begin{aligned} &u_{1}=a_{1} x+b_{1} y+c_{1}=0 \\ &u_{2}=a_{2} x+b_{2} y+c_{2}=0 \\ &u_{3}=a_{3} x+b_{2} y+c_{1}=0 \end{aligned} $$ $$ \text { is }\left|\begin{array}{ccc} \frac{1}{u_{1}} & \frac{1}{u_{2}} & \frac{1}{u_{3}} \\ a_{2} a_{3}-b_{2} b_{3} & a_{3} a_{1}-b_{3} b_{1} & a_{1} a_{2}-b_{1} b_{2} \\\ a_{2} b_{3}+a_{3} b_{2} & a_{3} b_{1}+a_{1} b_{3} & a_{1} b_{2}+a_{2} b_{1} \end{array}\right|=0 $$ $$ \text { or }\left|\begin{array}{ccc} \frac{a_{1}^{2}+b_{1}^{2}}{u_{1}} & \frac{a_{2}^{2}+b_{2}^{2}}{u_{2}} & \frac{a_{3}^{2}+b_{3}^{2}}{u_{3}} \\ a_{1} & a_{2} & a_{3} \\ b_{1} & b_{2} & b_{3} \end{array}\right|=0 $$

Expert verified

The proof is based on the principles of determinants and the condition for three lines to intersect at a single point. The given determinant forms are proven by substituting the given line equations, simplifying the determinants, and concluding that both equal zero when the lines are concurrent, thus forming a triangle with a well-defined circumcircle.

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