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
$x^2+y^2+ax+by+c=0$, then evaluate $\left|\frac{ab}{c}\right|.$

Answer 1

Answer: 0

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
$x^2+(y-2{)}^2=16$
$\iff x^2+y^2-4y-12=0$
$\Rightarrow a=0,b=-4,c=-12$
$\Rightarrow \left|\frac{ab}{c}\right|=0$