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  • The equation of the circumcircle of the triangle formed by the lines <img alt="\mathrm{y+\sqrt{3} x=6, y-\sqrt{3} x=6}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7By&plus;%

The equation of the circumcircle of the triangle formed by the lines \mathrm{y+\sqrt{3} x=6, y-\sqrt{3} x=6} and \mathrm{y=0} is

Option: 1

\mathrm{x^{2}+y^{2}-4 y=0}


Option: 2

\mathrm{x^{2}+y^{2}+4 x=0}


Option: 3

\mathrm{x}^{2}+\mathrm{y}^{2}-4 \mathrm{y}-12=0


Option: 4

\mathrm{x^{2}+y^{2}+4 x=12}


Answers (1)

best_answer

Solving the equations of lines in pairs, we obtain that the vertices of the \triangle \mathrm{ABC} are \mathrm{A}(0,6), \mathrm{b}$ $(-2 \sqrt{3}, 0) and \mathrm{C}(2 \sqrt{3}, 0).

\mathrm{Clearly \: \mathrm{AB}=\mathrm{BC}=\mathrm{CA}}.

\mathrm{So, \: \triangle \mathrm{ABC}} is an equilateral triangle. Therefore centroid of the triangle \mathrm{ABC} coincides with the circumcentre. Coordinates of the circumcentre are G(0,2) and the radius =G A=4.

Hence, the equation of the circumcircle is (\mathrm{x}-0)^{2}+(\mathrm{y}-2)^{2}=4^{2}$ or $\mathrm{x}^{2}+\mathrm{y}^{2}-4 \mathrm{y}=12.

Posted by

Ritika Jonwal

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