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

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

Correct Answer: C

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 \[=\frac{AC}{2\sin B}\] \[=\frac{1}{2}.\frac{2a}{\sqrt{3}}.\frac{2}{\sqrt{3}}=\frac{2a}{3}\] \[\therefore \]Required equation of circumcircle is \[{{x}^{2}}+{{y}^{2}}=\frac{4{{a}^{2}}}{9}\] \[\Rightarrow \]         \[9{{x}^{2}}+9{{y}^{2}}=4{{a}^{2}}\]