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  • If two vertices of an equilateral triangle, are  and <img alt="B(a,0)" src="http://entrancecor

If two vertices of an equilateral triangle, are A(-a,0) and B(a,0), a> 0, and the third vertex C lies above x-axis then the equation of the circumcircle of \triangle ABC is 

Option: 1

3x^{2}+3y^{2}-2\sqrt{3}ay=3a^{2}


Option: 2

3x^{2}+3y^{2}-2ay=3a^{2}


Option: 3

x^{2}+y^{2}-2ay=a^{2}


Option: 4

x^{2}+y^{2}-\sqrt{3}ay = a^{2}


Answers (1)

best_answer

We have given two vertices of an equilateral triangle, as A(-a,0) and B(a,0), a> 0

The midpoint of AB is (0, 0)

So, the third vertex of the triangle, C will lie on the y-axis. C= (0,k)

Coordinate of centroid, G of the triangle is

G=\left (\frac{a+(-a)+0}{3},\frac{0+0+k}{3} \right )=\left ( 0,\frac{k}{3} \right )

We also know that circumcenter, centroid, and incenter of the equilateral triangle coincides.  

\text { Hence, the coordinates of circumcenter is }\left(0, \frac{k}{3}\right) \text { . }

Since ABC is an equilateral triangle, So

\\\mathrm{AB}=\mathrm{AC}\\ \Rightarrow(\mathrm{AB})^{2}=(\mathrm{AC})^{2}\\ \Rightarrow \mathrm{k}=\sqrt{{3}}a\\ \text { So, the coordinates of circumcenter is }\left(0, \frac{a}{\sqrt{3}}\right) \text { . }\\ \text { Radius } \mathrm{CG}=\frac{2a}{\sqrt{3}}

\\\text{Equation of circumcircle is} \\ (x-0)^{2}+\left(y-\frac{a}{\sqrt{3}}\right)^{2}=\frac{4a^2}{3} \\\Rightarrow x^{2}+y^{2}-\frac{2a}{\sqrt{3}} y+\frac{a^2}{3}=\frac{4a^2}{3}\\ \Rightarrow {3}\left(x^{2}+y^{2}\right)-2\sqrt{3}a y-3a^2=0

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vinayak

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