Question

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

• A. $x^2+y^2+3x+3y-4=\pm \sqrt{3}(x+y-3)$
• B. $x^2+y^2-3x+3y=\pm \sqrt{3}(x+y-3)$
• C. $x^2+y^2-3x-3y+4=\pm \sqrt{3}(-x-y+3)$
• D. $x^2+y^2+3x-3y=\pm \sqrt{3}(x+y-3)$

Answer 1

The correct option is C $x^2+y^2-3x-3y+4=\pm \sqrt{3}(-x-y+3)$


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.
$\therefore \angle QPR={30}^{\circ }$
Since $\tan \theta =\left|\frac{m_2-m_1}{1+m_1{m}_2}\right|$
where $m_2$ is the slope of $PR$ and $m_1$ is the slope of $PQ,$
equation of circumcircle to the hexagon is
$\frac{\frac{y-2}{x-1}-\frac{y-1}{x-2}}{1+\frac{y-2}{x-1}\times \frac{y-1}{x-2}}=\pm \tan {30}^{\circ }$
$\Rightarrow (x-1)(x-2)+(y-1)(y-2)=\pm \sqrt{3}\left[\right(y-2\left)\right(x-2)-(x-1\left)\right(y-1\left)\right]$
$\Rightarrow x^2+y^2-3x-3y+4=\pm \sqrt{3}(-x-y+3)$