The equation of circle with origin as a centre and passing through equilateral triangle whose median is of length $3a$ ;is

Given, median of the equilateral triangle is $3a$ say $LD$

In $\displaystyle \Delta LMD$, we have

$(LM)^{2}=(LD)^{2}+(MD)^{2}$

$\Rightarrow \displaystyle (LM)^{2}=9a^{2}+\left(\frac{LM}{2}\right)^{2}$

$\displaystyle\Rightarrow \frac{3}{4}(LM)^{2}=9a^{2}$

$\Rightarrow (LM)^{2}=12a^{2}$

Again in triangle $\displaystyle OMD$,

$(OM)^{2}=(OD)^{2}+(MD)^{2}$

$\Rightarrow \displaystyle R^{2}=(3a-R)^{2}+ \left(\frac{LM}{2}\right)^{2}$

$\displaystyle\Rightarrow R^{2} =9a^{2}+R^{2}-6aR+3a^{2}$

$\displaystyle \Rightarrow 6aR=12a^{2}$

$\Rightarrow R=2a$

So, equation of circle is

$\displaystyle (x-0)^{2}+(y-0)^{2}=(2a)^{2}$

$\Rightarrow x^{2}+y^{2}=4a^{2}$