Question

If two vertices of an equilateral triangle are $(-1,0)$ and $(1,0),$ the equation of its circumcircle is

• A. $\sqrt{3}{x}^2+\sqrt{3}{y}^2+2y-\sqrt{3}=0$
• B. $2x^2+2y^2+\sqrt{3}y-2=0$
• C. $\sqrt{3}{x}^2+\sqrt{3}{y}^2-2y-\sqrt{3}=0$
• D. $2x^2+2y^2-\sqrt{3}y-2=0$

Answer 1

Answer: (A), (C)

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

Since two of the vertices $A(-1,0)$ and $B(1,0)$ lie on x-axis,

the third vertex $C$ lies on y-axis (figure)

Let the coordinate of $C$ be $(0,x)$
Then$(CA{)}^2=(AB{)}^2=(BC{)}^2=4.$
$\Rightarrow 1+x^2=4\Rightarrow x^2=3\Rightarrow x=\pm \sqrt{3}\Rightarrow OC=\pm \sqrt{3}$
Since triangle is equilateral, circurncentre of the triangle coincides with its centroid,

i.e. with $(0,\pm 1\sqrt{3})$ and circurnradius is $\frac{2}{\sqrt{3}}$ .

Hence possible equations of the circumcircle is
$x^2{(y\pm \frac{1}{\sqrt{3}})}^2={\left(\frac{2}{\sqrt{3}}\right)}^2$
$\Rightarrow x^2+y^2\pm \frac{2}{\sqrt{3}}y-1=0$
$\Rightarrow \sqrt{3}{x}^2+\sqrt{3}{y}^2\pm 2y-\sqrt{3}=0$