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  • 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   <img alt="\mathrm{x^2+y^2

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  

\mathrm{x^2+y^2+k \frac{2 x}{\sqrt{3}}+\ell \frac{2 y}{\sqrt{3}}-1=0}, where 

Option: 1

\mathrm{k=0, \ell=-1}


Option: 2

\mathrm{k=0, \ell=1}


Option: 3

\mathrm{k=1, \ell=0}


Option: 4

\mathrm{k=-1, \ell=0}


Answers (1)

best_answer

As D is mid point of BC, take DA as y-axis.
Let A be \mathrm{(0, \ell), B C=2 ; \ell>0} being above x-axis.

\mathrm{\begin{aligned} & \Rightarrow \sin 60=\frac{A D}{A B} \text { or } A D=2 \sin 60 \\ & \text { Or } A D=2 \cdot \frac{\sqrt{3}}{2}=\sqrt{3} \end{aligned}}

For an equilateral triangle, centroid \mathrm{G\left(0, \frac{\ell}{3}\right)=\left(0, \frac{\sqrt{3}}{3}\right)} 

is the circumcentre and the circum radius is \mathrm{A G=\frac{2}{3} \ell=\frac{2}{3} \sqrt{3}=\frac{2}{\sqrt{3}}} 

The equation of circumcircle is

\mathrm{ x^2+\left(y-\frac{1}{\sqrt{3}}\right)^2=\left(\frac{2}{\sqrt{3}}\right)^2}

\mathrm{ Or \: \: x^2+y^2-\frac{2 y}{\sqrt{3}}-1=0}

 

Posted by

mansi

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