A circle having its centre at the origin pass | vedclass

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Question

(C) Let the equilateral triangle be $	riangle ABC$ and its median be $AD = 9$ units.In an equilateral triangle,the centroid $O$ divides the median $A

Vedclass · Accepted Answer

$x^2+y^2=36$

Vedclass · Answer

(C) Let the equilateral triangle be $	riangle ABC$ and its median be $AD = 9$ units.In an equilateral triangle,the centroid $O$ divides the median $AD$ in the ratio $2:1$.Since the circle is centered at the origin $O(0,0)$ and passes through the vertices,$O$ is the circumcentre of $	riangle ABC$.In an equilateral triangle,the circumcentre and the centroid coincide.Therefore,the distance from the centre $O$ to the vertex $A$ is the radius $R$ of the circle.$R = AO = \frac{2}{3} AD = \frac{2}{3} 	imes 9 = 6$ units.The equation of a circle with centre at the origin $(0,0)$ and radius $R$ is $x^2 + y^2 = R^2$.Substituting $R = 6$,we get $x^2 + y^2 = 6^2 = 36$.

$A$ circle having its centre at the origin passes through the three vertices of an equilateral triangle,the length of whose median is $9$ units. Then,the equation of that circle is

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