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Two vertices of an equilateral triangle ...

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 ____________

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

Text Solution

Verified by Experts

The correct Answer is:
B
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`
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Knowledge Check

  • 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 circumcircle is

    A
    `x^(2)+y^(2)+(2x)/(sqrt(3))-1=0`
    B
    `x^(2)+y^(2)-(2x)/(sqrt(3))-1=0`
    C
    `x^(2)+y^(2)+(2y)/(sqrt(3))-1=0`
    D
    `x^(2)+y^(2) -(2y)/(sqrt(3))-1=0`

  • If two vertices of an equilateral triangle are (1,1) and (-1,1) then the third vertex may be

    A
    `(-sqrt3,-sqrt3)`
    B
    `(-sqrt3,sqrt3)`
    C
    `(sqrt3,-sqrt3)`
    D
    `(sqrt3,sqrt3)`

  • If two vertices of an equilateral triangle are A (-a,0)and B (a,0), a gt 0, and the third vertex C lies above y-axis, then the equation of the circumcircle of Delta ABC is :

    A
    `3x ^(2) + 3y^(2) - 2 sqrt3 ay = 3a ^(2)`
    B
    `3x ^(2) + 3y ^(2) - 2ay = 3a ^(2)`
    C
    `x ^(2) +y ^(2) - 2ay =a ^(2)`
    D
    `x ^(2) +y ^(2) - sqrt3 ay =a ^(3)`

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