Question

The equation of the circumcircle of the triangle formed by the lines $y+\sqrt{3} x=6, y-\sqrt{3} x=6$ and $y=0$ is
$x^{2}+y^{2}+a x+b y+c=0$, then evaluate $\left|\frac{a b}{c}\right| .$

Answer 1

Let $l_{1}: y+\sqrt{3} x=6$
$l_{2}: y-\sqrt{3} x=6$
$l_{3}: y=0$
Solving the equations of lines $l_{1}, l_{2}, l_{3}$ pairwise, we get
$A=(0,6), B=(-2 \sqrt{3}, 0)$
and $C =(2 \sqrt{3}, 0)$
$\Rightarrow| AB |=| BC |=| CA |=4 \sqrt{3}$
$\Rightarrow \Delta$ is an equilateral triangle.
$\Rightarrow$ The centroid and the circumcentre are same.
$\Rightarrow $ Circumcentre $=\left(\frac{0+2 \sqrt{3}-2 \sqrt{3}}{3}, \frac{6+0+0}{3}\right) $
$=(0,2)$
Circumradius $=\frac{2}{\sqrt{3}}\left(\frac{4 \sqrt{3}}{2}\right)=4$
Equation of circumcircle is
$x^{2}+(y-2)^{2}=16$
$\Leftrightarrow x^{2}+y^{2}-4 y-12=0$
$\Rightarrow a=0, b=-4, c=-12$
$\Rightarrow \left|\frac{a b}{c}\right|=0$