question_answer

An equilateral triangle is inscribed in the circle \[{{x}^{2}}+{{y}^{2}}={{a}^{2}}\]with one of the vertices at (a, 0). What is the equation of the side opposite to this vertex?

A) \[2x-a=0\]

B) \[x+a=0\]

C) \[2x+a=0\]

D) \[3x-2a=0\]

Correct Answer: C

Solution :

[c] Since the equilateral triangle is inscribed in the circle with centre at the origin, centroid lies on the origin.

So, \[\frac{AO}{OD}=\frac{2}{1}\]

and \[OD=\frac{1}{2}AO=\frac{a}{2}\]

So, other vertices of triangle have coordinates,

\[\left( -\frac{a}{2},\frac{\sqrt{3a}}{2} \right)\] and \[\left[ -\frac{a}{2},-\frac{\sqrt{3}}{2}a \right]\]

\[\left( -\frac{a}{2}\frac{\sqrt{3a}}{2} \right)y\]

\[\left( -\frac{a}{2},\frac{-\sqrt{3a}}{2} \right)\]

\[\therefore \] Equation of line BC is:

\[x=-\frac{a}{2}\]

\[\Rightarrow \,\,\,\,\,2x+a=0\]