Question

Two vertices of an equilateral triangle are (-1,0) and (1,0) and its third vertex lie above the x axis , an equation of the circumcircle is
1) 3x​2+3y2-2√3y = 3
2) 2x2+y2-3√3y = 0
3) x2+y2-2y = 1
4) none of these

Answer 1

It has been given that it is an equilateral triangle.

Let the coordinates are be $A(-1,0)$ and $B(1,0)$ respectively. Since the third vertex lies above x-axis and y-axis is the perpendicular bisector of AB.

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

Length of the side AB is $1+1=2$ units.

Since it is an equilateral triangle therefore, the length of the third side that is AC must be 2 units.

Now,

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

$p^2+1=4$

$p^2=3$

$p= \sqrt 3$

Hence, the coordinates of point C is $(0,\sqrt3)$

Let the equation of the required circle be

$x^2+y^2+2gx+2fy+c=0$

Since it passes through $(-1,0)$

Therefore, $1-2g+c=0$............(1)

and it also passes through $(1,0)$

$1+2g+c=0$.........(2)

Solving equation 1 and 2 we get:

$-4g=0$

$g=0$

and $2+2c=0$

$c=-1$

Now, the circle also passes through the point $(0,\sqrt3)$

$3+2\sqrt3f+c=0$..........(3)

Plugging the value of C we get in equation 3:

$3+2\sqrt3f-1=0$

$f=\frac{-1}{\sqrt3}$

So the equation of the required circle is:

$x^2+y^2-\frac{2}{\sqrt3} y-1=0$

multiplying the equation by 3 we get:

$3x^2+3y^2-2\sqrt3y=3$