Let the circle with centre at origin pass through the vertices of an equilateral triangle $A B C$. If $A \equiv(2,4)$, then the length of the median through A is

Given the circle centered at $O(0,0)$ passes through the vertices of equilateral triangle ABC, with vertex $A$ at $(2,4)$.

For an equilateral triangle, the centroid $G$ coincides with the circumcenter, hence $G = (0,0)$.

The centroid divides the median $AM$ in the ratio $2:1$, so $AG = \frac{2}{3} AM$. The distance between $A(2,4)$ and $G(0,0)$ is $AG = \sqrt{(2-0)^2 + (4-0)^2} = \sqrt{4 + 16} = \sqrt{20} = 2\sqrt{5}$.

Substituting into the centroid relation, $2\sqrt{5} = \frac{2}{3} AM$, so $AM = \frac{3}{2} \cdot 2\sqrt{5} = 3\sqrt{5}$.

Thus, the length of the median through $A$ is $3\sqrt{5}$ units, corresponding to option B.