Question

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. $3x^2+3y^2=a^2$
• E. $9x^2+9y^2=4a^2$

Answer 1

Answer: (E)

Centre of triangle is (0, 0).
$\therefore$ Since triangle is an equilateral, the centre of circumcircle is also (0, 0) $AD=a$ (given)
$\therefore$ $AC=BC=AB=\frac{a}{\sin 60{}^{\circ }}=\frac{2a}{\sqrt{3}}$

$\therefore$ Circumradius
$=\frac{AC}{2\sin B}$
$=\frac{2a}{\sqrt{3}.2}\times \frac{2}{\sqrt{3}}$ $(\because B=60{}^{\circ })$
$=\frac{2a}{3}$
$\therefore$ Required equation of circumcircle is $x^2+y^2=\frac{4a^2}{9}\Rightarrow 9x^2+9y^2=4a^2$