Find the equation of the circle with origin as center and passing through the vertices of an equilateral triangle whose median is of length $3a$.
In an equilateral triangle
- the medians bisect the side which they meet at right angles.
- the medians intersect at a point which is the circumcenter of the circumscribed circle.
- the medians intersect at a point (centroid) which is $\dfrac{2}{3}^{rd}$ the length of the median, measured from the respective vertex.
Let $ABD$ be an equilateral triangle.
Let $DM$ be the median of the triangle.
Let $C$ be the centroid.
Then $C$ is the center of the circumcircle.
Given: $C = \left(0,0\right)$; $\;\;$ length of median $= DM = 3a$
$\therefore$ $\;$ $DC = \dfrac{2}{3} \times DM = \dfrac{2}{3} \times 3a = 2a$
But radius of required circle $= DC$
$\therefore$ $\;$ Equation of required circle is
$x^2 + y^2 = \left(2a\right)^2$
i.e. $x^2 + y^2 = 4a^2$