Question

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

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

Answer 1

The correct option is B $x^2+y^2+\frac{2y}{\sqrt{3}}-1=0$

Let coordinates be $A(-1,0)$ and $B(1,0)$ respectively ,

Since third vertex lies above $x$ axis

and $y$ is $\perp$ of $AB$

Let third coordinate be $C(0,p)$

Since, the third vertex will have $x-coordinate=0$

Length of $AB=1+1=2$ units

Since $△$ is equilateral

Therefore, $AC=2$ units,

New, $\sqrt{1^2+p^2}=2$

$p^2=4-1$

$p=\sqrt{3}$

$C=(0,\sqrt{3})$

Let equation of circle $:x^2+y^2+2gx+2fy+c=0$

passes through $(-1,0)$

Therefore, $-2g+c=0------\left(1\right)$

also passes through $(1,0)$

$1+2g+c=0------\left(2\right)$ from $eq\left(1\right)$ and $\left(2\right)$

$g=0,c=-1$

Circle passes through $(0,\sqrt{3})$

$3+2\sqrt{3}f+c=0-------\left(3\right)$

Putting value of $c$ in $eq\left(3\right)$

$f=-1/\sqrt{3}$

$E{q}^n$ of circle

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

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