Question

The equation of circle with origin as center 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. $x^2+y^2=a^2$

Answer 1

Answer: (C)

$x^2+y^2=4a^2$

The centroid of an equilateral triangle is the center of its circumcenter and the radius of the circle is the distance of any vertex from the centroid i.e. radius of the circle

$=$ distance of centroid from any vertex

$=\frac{2}{3}($Median$)=\frac{2}{3}\left(3a\right)=2a$

Hence, equation of circle whose center is $(0,0)$ and radius $2a$ is

${(x-0)}^2+{(y-0)}^2={\left(2a\right)}^2\Rightarrow x^2+y^2=4a^2$