question_answer

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-

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

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

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

D) None of these

Correct Answer: B

Solution :

[b] Third vertex is \[\left( \frac{{{x}_{1}}+{{x}_{2}}\pm \sqrt{3}({{y}_{1}}-{{y}_{2}})}{2},\frac{{{y}_{1}}+{{y}_{2}}\mp \sqrt{3}({{x}_{1}}-{{x}_{2}})}{2} \right)\] \[\left( \frac{-1+1\pm \sqrt{3}(0-0)}{2},\frac{0+0\mp \sqrt{3}(-1-1)}{2} \right)\] \[=\left( 0,\pm \sqrt{3} \right)\] But third vertex lies above x-axis \[\therefore \] it will be \[(0,\sqrt{3})\] Triangle is equilateral \[\therefore \] circumcentre = centroid \[G\equiv \left( 0,\frac{1}{\sqrt{3}} \right)\]  circumradius \[GA=\frac{2}{\sqrt{3}}\] \[\therefore \] Equation of circumcircle \[\Rightarrow {{(x-0)}^{2}}+{{\left( y-\frac{1}{\sqrt{3}} \right)}^{2}}=\frac{4}{3}\] \[\Rightarrow {{x}^{2}}+{{y}^{2}}-\frac{2}{\sqrt{3}}y-1=0\]