CircleEasy
Question
The equation of a circle with origin as a centre and passing through equilateral triangle whose median is of length 3a is
Options
A.x2 + y2 = 9a2
B.x2 + y2 = 16a2
C.x2 + y2 = 4a2
D.x2 + y2 = a2
Solution
Let ABC be an equilateral triangle, whose median is AD.
Given AD = 3a
In ᐃABD, AB2 = AD2 + BD2 ;
⇒ x2 = 9a2 + (x2/4) whereAB = BC = AC = x.
x2 = 9a2 ⇒ x2 = 12a2In ᐃOBD, OB2 = OD2 + BD2
⇒ r2 = (3a - r)2 +
⇒ r2 = 9a2 - 6ar + r2 + 3a2 ; ⇒ 6ar = 12a2 ⇒ r = 2a So equation of circle is x2 + y2 = 4a2
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