Question

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

Answer 1

Consider $A(-1,0)$ and $B(1,0)$

let $AB$ be base of the Triangle

The third point is on $y-axis$ at a height of $\frac{\sqrt{3}}{2}\times 2$ i.e.,$C(0,\sqrt{3})$

General Equation of the circle is $x^2+y^2+2gx+2fy+c=0$

Circle passes through $A(-1,0)$

$⟹1-2g+c=0\cdots \left(1\right)$

Circle passes through $B(1,0)$

$⟹1+2g+c=0\cdots \left(2\right)$

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

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

From $\left(1\right),\left(2\right),\left(3\right)$

$⟹c=-1,g=0,f=\frac{-1}{\sqrt{3}}$

$⟹$ Equation of circle is $\sqrt{3}(x^2+y^2)-2y-\sqrt{3}=0$