Let ABC be an equilateral triangle, whose median is AD. In equilateral triangle median is also altitude So, AD⟂BC Given AD=3a Let AB=BC=AC=x. In △ABD,AB2=AD2+BD2 ⇒x2=9a2+(x2∕4)
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x2=9a2⇒x2=12a2 In ∆OBD,OB2=OD2+BD2 ⇒r2=(3a−r)2+
x2
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⇒r2=9a2−6ar+r2+3a2 ⇒6ar=12a2 ⇒r=2a So equation of circle is x2+y2=4a2