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  • Let <img alt="\mathrm{A(-2,0), B(2,0) \: and \: C\left(x_1, y_1\right)\: where\: y_1 \in R^{+}}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7BA%28-2%2C0%29%2C%20B%282%2C0%29%20%

Let \mathrm{A(-2,0), B(2,0) \: and \: C\left(x_1, y_1\right)\: where\: y_1 \in R^{+}}be the vertices of an equilateral triangle \mathrm{A B C}. If \mathrm{D} be any point such that \mathrm{\angle \mathrm{BDC}=\frac{\pi}{2}}, then the equation of the circumcircle of \mathrm{\triangle \mathrm{BDC}} is
 

Option: 1

\mathrm{x^2+y^2-\sqrt{3} x-2 y=0}

 


Option: 2

\mathrm{ x^2+y^2-2 x-2 \sqrt{3} y=0}
 


Option: 3

\mathrm{x^2+y^2-2 x-\sqrt{3} y+1=0}
 


Option: 4

\mathrm{None \: of \: these}


Answers (1)

best_answer

Clearly \mathrm{x}_1=0, \mathrm{y}_1=2 \sqrt{3}

\therefore equation of circumcircle of \mathrm{B D C=x(x-2)+y(y-2 \sqrt{3})=0}

Hence option 2 is correct.

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Gunjita

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