Q. For an equilateral triangle the centre is the origin and the length of altitude is$a$. Then, the equation of the circumcircle is
A) $x^{2} + y^{2} = a^{2}$
B) $3x^{2} + 3y^{2} = 2a^{2}$
C) $x^{2} + y^{2} = 4a^{2}$
D) $9x^{2} + 9y^{2} = 4a^{2}$
Explanation
(d)
Centre of triangle is (0, 0)
Since, triangle is an equilateral, the centre of circumcircle is also (0, 0)
$AD = a$ (given)
[image:11_image73.emf]
$\therefore\ AC = BC = AB$
$= \frac{a}{\sin 60{^\circ}} = \frac{2a}{\sqrt{3}}$
$\therefore$ Circumradius $= \frac{AC}{2\sin B}$
$= \frac{2a}{2\sqrt{3}} \times \frac{2}{\sqrt{3}} = \frac{2a}{3}\ \ \ \ \ \lbrack\because B = 60{^\circ}\rbrack$
$\therefore$ required equation of circumcircle is
$x^{2} + y^{2} = \frac{4a^{2}}{9}$
$\Rightarrow \ 9x^{2} + 9y^{2} = 4a^{2}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.