Origin is centre of circle passing through the vertices of an equilateral triangle whose median is of length 3a then equation of circle is

Given that equilateral triangle is inscibed in a circle  then

Centre of circle == Centroid of triangle

$$\begin{gathered} So,centre of circle \left(C\right)=(0,0) \\ and length of median =3a \\ Now radius of circle\left(r\right)=\frac{2}{3}\left(3a\right)=2a \end{gathered}$$

$$\begin{gathered} \Rightarrow \mathrm{r}=2\mathrm{a} \\ Now required circle is \\ {\mathrm{x}}^2+{\mathrm{y}}^2={\left(2\mathrm{a}\right)}^2 \\ \Rightarrow {\mathrm{x}}^2+{\mathrm{y}}^2=4{\mathrm{a}}^2 \end{gathered}$$