Question

Two vertices of an equilateral triangle are (–1, 0) and (1, 0) and its third vertex lies above the x-axis, the equation of the circumcircle is

• A. $3x^2+3y^2-2\sqrt{3}y=3$
• B. $2x^2+2y^2-3\sqrt{2}y=3$
• C. $x^2+y^2-2y=1$
• D. $None$

Answer 1

The correct option is A $3x^2+3y^2-2\sqrt{3}y=3$

Let A(–1, 0), B(1, 0) and C(0, b) be the vertices of the triangle, since C lies on the locus of points equidistance from A(–1, 0), and B(1, 0) i.e., y-axis.
Then $AB=AC$
$\Rightarrow \sqrt{1+b^2}=2$
$\Rightarrow b^2=3$
$\Rightarrow b=\sqrt{3}$ [$\because$ b>0]
Since the triangle is equilateral, the centre of the circumcircle is at the centroid of the triangle which is $(0,\frac{1}{\sqrt{3}})$.
Thus the equation of the circumcenter is,
$(x-0{)}^2+{(y-\frac{1}{\sqrt{3}})}^2=(1-0)2+{(0-\frac{1}{\sqrt{3}})}^2$
$\Rightarrow x^2+y^2-\left(\frac{2}{\sqrt{3}}\right)y+\frac{1}{3}=1+\frac{1}{3}$
$\Rightarrow 3x^2+3y^2-2\sqrt{3}y=3$