The equation of a circle with origin as a centre and passing...

Let $A B C$ be an equilateral triangle, whose median is $A D$. Given $A D=3 a$ In $\triangle \mathrm{ABD}, \mathrm{AB}^2=\mathrm{AD}^2+\mathrm{BD}^2$ $\Rightarrow x^2=9 a^2+$\left(x^2 / 4\right)$where$A B=B C=A C=x . \frac{3}{4} x^2=9 a^2 \Rightarrow x^2=12 a^2$In$\triangle \mathrm{OBD}, \mathrm{OB}^2=\mathrm{OD}^2+\mathrm{BD}^2 \Rightarrow r^2=(3 a-r)^2+\frac{x^2}{4} \Rightarrow r^2=9 a^2-6 a r+r^2+3 a^2 ; \Rightarrow 6 a r=12 a^2 \Rightarrow r=2 a$So equation of circle is$x^2+y^2=4 a^2