A circle circumscribing an equilateral triangle with centroid at (0,0) of side ‘a’ is drawn and a square is drawn whose four sides touch the circle. The equation of the circle circumscribing the square is

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$5 x^{2} + 5 y^{2} = 3 a^{2}$

$x^{2} + y^{2} = \frac{a^{2}}{2}$

$3 x^{2} + 3 y^{2} = 2 a^{2}$

$x^{2} + y^{2} = 2 a^{2}$

answer is B.

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Detailed Solution

Clearly, centre of the circum circle is at origin and radius is two-third of the altitude $\frac{\sqrt{3}}{2} a$ .

So, the equation of the circumcircle is $x^{2} + y^{2} = {(\frac{a}{\sqrt{3}})}^{2}$
Clearly the radius of the circumcircle of the square ABCD $= O A = \sqrt{O L^{2} + A L^{2}} = \sqrt{\frac{a^{2}}{3} + \frac{a^{2}}{3}} = \sqrt{\frac{2}{3}} a$
So, the equation of the circumcircle is $x^{2} + y^{2} = \frac{2}{3} a^{2} \Rightarrow 3 (x^{2} + y^{2}) = 2 a^{2}$