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

• 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)

We know that centroid divides the median in the ratio $2:1$
Radius of the circle $=\frac{2}{3}\times $ length of median
$=\frac{2}{3}\times 3a=2a$
Centre of the (given) circle is $C\left(0 , 0\right)$
Therefore the equation of the circle
$\left(x - 0\right)^{2}+\left(y - 0\right)^{2}=\left(2 a\right)^{2}\Rightarrow x^{2}+y^{2}=4a^{2}$