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# 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

[Hint : Centroid of the triangle coincides with the centre of the circle and the radius of the circle is $\large\frac{2}{3}$ of the length of the median]

$\begin{array}{1 1}x^2+y^2=9a^2\\x^2+y^2=16a^2\\x^2+y^2=4a^2\\x^2+y^2=a^2\end{array}$

• Equation of a circle passing through the origin is $x^2+y^2=a^2$
Answer : $x^2+y^2=4a^2$
If the centroid of the triangle coincides with the centre of the circle,then the radius of the circle is $\large\frac{2}{3}$ of the length of the median.
$\therefore r=\large\frac{2}{3}$$\times 3a$
$\Rightarrow r^2=4a^2$
Hence the equation of the circle is $x^2+y^2=4a^2$