Question

If two adjacent vertices of a regular hexagon are $(0,0)$ and $(1,2),$ then equation of the circumcircle of the hexagon is

• A. $x^2+y^2-x-2y=\pm 3$
• B. $x^2+y^2-x-2y=\pm \sqrt{3}$
• C. $x^2+y^2-x-2y=\pm \sqrt{3}(2x-y)$
• D. $x^2+y^2-x-2y=0$

Answer 1

The correct option is C $x^2+y^2-x-2y=\pm \sqrt{3}(2x-y)$


We know that the measure of the angle formed by each side of a regular hexagon at the centre is ${60}^{\circ }.$
Also, the angle subtended by an arc at the centre of a circle is twice the angle subtended at its circumference.
$\therefore \angle OPQ={30}^{\circ }$
Since $\tan \theta =\left|\frac{m_2-m_1}{1+m_1{m}_2}\right|$
where $m_2$ is the slope of $OP$ and $m_1$ is the slope of $PQ,$
$\therefore$ Equation of circumcircle of the hexagon is
$\frac{\frac{y-0}{x-0}-\frac{y-2}{x-1}}{1+\frac{y-0}{x-0}\times \frac{y-2}{x-1}}=\pm \tan {30}^{\circ }$

$\Rightarrow \frac{y(x-1)-x(y-2)}{x(x-1)+y(y-2)}=\pm \frac{1}{\sqrt{3}}$
$\Rightarrow x^2+y^2-x-2y=\pm \sqrt{3}(2x-y)$