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 \[{{x}^{2}}+{{y}^{2}}=\frac{4{{a}^{2}}}{9}\Rightarrow 9{{x}^{2}}+9{{y}^{2}}=4{{a}^{2}}\]You need to login to perform this action.
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