Q50 of 63 Page 202

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


In triangle LMD,


(LM)2 = (LD)2 + (MD)2 - Pythagoras Theorem,


LD = 3a (Given) &





(LM)2 = 12a2


In triangle OMD,


(OM)2 = (OD)2 + (MD)2 - Pythagoras Theorem,



As LO + OD = LD = 3a,



[Putting (LM)2 = 12a2]


9a2 + 3a2 - 6aR = 0


6aR = 12a2


R = 2a


Since, the equation of a circle having centre (h,k), having radius as "r" units, is


(x – h)2 + (y – k)2 = r2


(x – 0)2 + (y – 0)2 = (2a)2


x2 + y2 = 4a2


Hence the required equation is x2 + y2 = 4a �.


Option (D) is the answer.

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