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Figure shows a circle whose origin is at center passes through an equilateral triangle ABC .

Median of triangle is AD = ( 3 a ) and median passes through origin .

ΔBAD is right triangle having angles 90°, 60° and 30°.

Hence if AD = 3a , then AB = ( 2√3 a ) .............................(1)

( use the property of right triangle having angles 90°, 60° and 30° have sides in the ratio 2 : √3 : 1 ) .

In the ΔAOB , ∠AOB = 120° . Let OA = OB = r is radius of circle.

Then by cosine rule ,

Using eqn.(1) and above expression , we get

√3 r = 2 √3 a

Hence radius r of circle is 2a .

If centre of circle is at origin and radius is 2a , then

x² + y² = (2a)² = 4a²