Question

Vertices of an equilateral triangle are $(-1,0)$ and $(1,0)$ and its third vertex lies above the x-axis an equation of the circumcircle is ?

• A. $\begin{array}{c}{x}^2+y^2-\frac{1}{\sqrt{3}}y-1=0\end{array}$
• B. $x^2+y^2+\frac{2}{\sqrt{3}}y-1=0$
• C. $\begin{array}{c}{x}^2+y^2-\frac{2}{\sqrt{3}}y-1=0\end{array}$
• D. none of these

Answer 1

Answer: (C)

$\begin{array}{c}{x}^2+y^2-\frac{2}{\sqrt{3}}y-1=0\end{array}$

It is given that $△ABC$ is an equilateral triangle, where the co-ordinates are $A(-1,0)$ and $B(1,0)$ respectively and the third vertex , say C, lies above the X-axis, C will lie on the Y-axis, as Y-axis is the perpendicular bisector of the line segment $AB$.

Let the co-ordinates of C be $(0,k)$.

Length of side $AB=2units$.

$\therefore$ Length of $AC$ should also be $2units$.

$\Rightarrow \sqrt{1^2+p^2}=2$

$\Rightarrow 1^2+p^2=4$

$\Rightarrow p=\pm \sqrt{3}$

But since, C lies above the X-axis we will have its co-ordinates as $(0,\sqrt{3})$.

Now, the circumference of the above triangle will have its center at the centroid of the $△ABC$ (Since its an equilateral triangle, the circumcenter, incenter & centroid coincide).

$\therefore$ Center of circle $C\equiv (\frac{-1+1+0}{3},\frac{0+0+\sqrt{3}}{3})$

$=(0,\frac{1}{\sqrt{3}})$

and radius is $\sqrt{1^2+(\frac{1}{\sqrt{3}}{)}^2}=\sqrt{1+\frac{1}{3}}=\frac{2}{\sqrt{3}}$

$\therefore$ Equation of circle is

$(X-0{)}^2+(Y-\frac{1}{\sqrt{3}}{)}^2=\frac{4}{3}$

$\Rightarrow X^2+Y^2+\frac{1}{3}-\frac{2Y}{\sqrt{3}}=\frac{4}{3}$

$\Rightarrow X^2+Y^2-\frac{2Y}{\sqrt{3}}-1=0$