Question
Two vertices of an equilateral triangle are (–1, 0) and (1, 0) and its third vertex lies above the x-axis, the equation of the circumcircle is
Answer: Option A
:
A
Let A(–1, 0), B(1, 0) and C(0, b) be the vertices of the triangle, since C lies on the locus of points equidistance from A(–1, 0), and B(1, 0) i.e., y-axis.
Then AB=AC
⇒√1+b2=2
⇒b2=3
⇒b=√3 [∵ b>0]
Since the triangle is equilateral, the centre of the circumcircle is at the centroid of the triangle which is (0,1√3).
Thus the equation of the circumcenter is,
(x−0)2+(y−1√3)2=(1−0)2+(0−1√3)2
⇒x2+y2−(2√3)y+13=1+13
⇒3x2+3y2−2√3y=3
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:
A
Let A(–1, 0), B(1, 0) and C(0, b) be the vertices of the triangle, since C lies on the locus of points equidistance from A(–1, 0), and B(1, 0) i.e., y-axis.
Then AB=AC
⇒√1+b2=2
⇒b2=3
⇒b=√3 [∵ b>0]
Since the triangle is equilateral, the centre of the circumcircle is at the centroid of the triangle which is (0,1√3).
Thus the equation of the circumcenter is,
(x−0)2+(y−1√3)2=(1−0)2+(0−1√3)2
⇒x2+y2−(2√3)y+13=1+13
⇒3x2+3y2−2√3y=3
Was this answer helpful ?
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