Two vertices of an equilateral triangle are `(-1,0)` and (1, 0), and its third vertex lies above the x-axis. The equation of its circumcircel is ____________

Options

A. `x^(2)+y^(2)-(2)/(sqrt(3))x-1=0`
B. `x^(2)+y^(2)-(2)/(sqrt(3))y-1=0`
C. `x^(2)+y^(2)+(2)/(sqrt(3))y-1=0`
D. None of these

Answer

B

Explanation

Third vertex is
`((x_(1)+x_(2)pmsqrt(3)(y_(1)-y_(2)))/(2),(y_(1)+y_(2)pmsqrt(3)(x_(1)-x_(2)))/(2))` ltBrgt `((-1+1pmsqrt(3)(0-0))/(2),(0+0pmsqrt(3)(-1-1))/(2))`
`(0,pmsqrt(3))` ltBrgt But third vertex lies above x-axis ltBrgt `therefore` it willl be (0, `sqrt(3)`) ltBrgt
Triangle is equilateral
`therefore` circumcentre `-=` centroid
`G-=(0,(1)/(sqrt(3)))` circumradius GA = `(2)/(sqrt(3))`
`therefore` Equation of circumcircle
`rArr (x-0)^(2)+(y-(1)/(sqrt(3)))^(2)=(4)/(3)` ltBrgt `rArrx^(2)+y^(2)-(2)/(sqrt(3))y-1=0`