The equation of a circle with origin as centre and passing through the vertices of an equi…

Question

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

Accepted Answer

In an equilateral triangle, the centroid (here the origin, which is also the circumcentre) divides each median in ratio $2:1$ from the vertex. Since the median has length $3a$, the circumradius is
$$r=\frac23\times3a=2a.$$
Circle: $x^2+y^2=4a^2.$

*(Verification via the median-altitude relation: with side $x$, $\tfrac34x^2=(3a)^2\Rightarrow x^2=12a^2$; the circumradius $r$ satisfies $r^2=(3a-r)^2+\tfrac{x^2}{4}$, giving $6ar=12a^2\Rightarrow r=2a$, confirming the same result.)*

Solutions · 1