Question

The equation of circumcircle of the triangle formed by the lines $y+\sqrt{3}x=6,y-\sqrt{3}x=6$ and $y=0$, is

• A. $x^2+y^2+4x=0$
• B. $x^2+y^2-4x=0$
• C. $x^2+y^2-4y=12$
• D. $x^2+y^2+4y=12$

Answer 1

Answer: (C)

$x^2+y^2-4y=12$

$y+\sqrt{3}x=6$, $y-\sqrt{3}x=6,y=0$

Intersection of $y+\sqrt{3}=6$ and $y-\sqrt{3}x=6$

$\Rightarrow$ $6+\sqrt{3}+\sqrt{3}=6$

$\Rightarrow$ $x=0$ $\Rightarrow$ $y=6$

Intersection of $y+\sqrt{3}=6$ and $y=0$

$\Rightarrow$ $x,y=0$

Intersection of $y-\sqrt{3}x+6=0$ and $y=0$

$x=-2\sqrt{3};y=0$

Slope of $y+\sqrt{3}x=6$ $\Rightarrow$ $m_1=-\sqrt{3}$

${\theta }_1=-{60}^o$

Slope of $y-\sqrt{3}x=6\Rightarrow$ $m_2=\sqrt{3}$

${\theta }_2={60}^o$

${\theta }_3=180-(60+60)={60}^o$

$\therefore$ $ABC$ is equilateral triangle

length $=4\sqrt{3}$

Circum center $=$ centroid $=(0,2)$

Radius $=\frac{length}{\sqrt{2}}=4$ [$r\cos \theta =\frac{a}{2}\Rightarrow \frac{\sqrt{3}r}{2}=\frac{a}{2}\Rightarrow r=\frac{a}{\sqrt{3}}$]

${(x-0)}^2+{(y-2)}^2=r^2$

$x^2+y^2+4-4y=16$

$\Rightarrow$ $x^2+y^2-4y=12$