Question

The equation of the circumcircle of the regular hexagon whose two consecutive vertices have the coordinates $(-1,0)$ and $(1,0)$ and which lies wholly above the $x$- axis, is

• A. $x^2+y^2-2\sqrt{2}y-1=0$
• B. $x^2+y^2-\sqrt{2}y-1=0$
• C. $x^2+y^2-2\sqrt{2}x-1=0$
• D. none of these

Answer 1

Answer: (D)

none of these

Solving it graphically, from the figure,
the radius of the circumscribing circle = 2
(Since all the triangle formed are equilateral)
Center of circle = $(0,2cos{30}^o)=(0,\sqrt{3})$
Equation of circle
$(x-0{)}^2+(y-\sqrt{3}{)}^2=4$