Q.The equation of a circle with the origin as centre and passing through the vertices of an equilateral triangle whose median is of length $3a$ is (A) $x^2+y^2=9a^2$ (B) $x^2+y^2=16a^2$ (C) $x^2+y^2=4a^2$ (D) $x^2+y^2=a^2$

For an equilateral triangle, the centroid and the circumcentre are the same point, and the centroid divides each median in the ratio $2:1$ from the vertex — so the circumradius (distance from centre to a vertex) is $\tfrac23$ of the median length. With median $=3a$, circumradius $=\dfrac2 …

Use the equilateral-triangle fact that the circumradius is $2/3$ of the median (since centroid = c …

• Using $\tfrac13$ of the median instead of $\tfrac23$ (confusing the two segments the centroid splits a median into) …