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

Step-by-step solution

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