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

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