Question

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

Answer 1

Two vertices of an equation triangle as $A=(-1,0),B=(1,0)$

third vertex ${}^{\prime }{C}^{\prime }$ lies above the $x-$axis

also given, to find the equation of the circum circle

Let us assume the third coordinate point $C=(0,r)$

length of the side $AB=1+1=2$ units

as it is the equation triangle, the length of the third side i.e., $AC=2$

Now, let us find ${}^{\prime }{r}^{\prime }$ point distance between $AC$ is $\sqrt{(x_2-x_1{)}^2+(y_2-y_1{)}^2}$

$(0,r)(-1,0)$

$x_1{y}_1{x}_2{y}_2$

$=\sqrt{(0-1{)}^2+(r-0{)}^2}\Rightarrow \sqrt{1+r^2}=AC$

$\Rightarrow r^2=3\iff r=\sqrt{3}$

$\therefore$ the co-ordinates of $C=(0,\sqrt{3})$

Let us assume equation of the circle as $x^2+y^2+2gx+2fy+c=0$ as it passes through $A(-1,0)$ we get,

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

and when circle equation passes through $B(1,0)$

we get, $1+2g+c=0\to eq\left(2\right)$

by solving equation $\left(1\right)$ and equation $\left(2\right)$ we get,

$\frac{1-2g+c=01+2g+c=0}{2+2c=0}\Rightarrow 2(1+c)=0\Rightarrow c=-1$

Let us substitute ${}^{\prime }{c}^{\prime }$ in equation $\left(1\right)\Rightarrow 1-2g+(-1)=0$

$\Rightarrow -2g=0\Rightarrow g=0$

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

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

as we got $c=-1$, let us substitute in equation $\left(3\right)$

$\Rightarrow 3+2\sqrt{3}f+(-1)=0\iff 2+2\sqrt{3}=0$

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

Now, let us substitude $c,g,f$ values in circle

equation, Hence we get

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

$\Rightarrow x^2+y^2-\frac{2}{\sqrt{3}}y-1=0\to eq\left(4\right)$

Multiply equation $\left(4\right)$ with ${}^{\prime }{3}^{\prime }$ we get,

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

$\Rightarrow 3x^2+3y^2-3-\frac{2}{\sqrt{3}}y-3=0[\because 3=\sqrt{3}.\sqrt{3}]$

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

$\therefore$ the equation of the circumcircle is $3x^2+3y^2-2\sqrt{3}y=3$