question_answer

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)  \[3{{x}^{2}}+3{{y}^{2}}=2{{a}^{2}}\]

C)  \[{{x}^{2}}+{{y}^{2}}=4{{a}^{2}}\]

D)  \[3{{x}^{2}}+3{{y}^{2}}={{a}^{2}}\]

E)  \[9{{x}^{2}}+9{{y}^{2}}=4{{a}^{2}}\]

Correct Answer: E

Solution :

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                 \[{{x}^{2}}+{{y}^{2}}=\frac{4{{a}^{2}}}{9}\Rightarrow 9{{x}^{2}}+9{{y}^{2}}=4{{a}^{2}}\]