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The equation of a circle with origin as a centre and passing through equilateral triangle whose median is of length 3a is
  • a)
    x2+y2=9a2
  • b)
    x2+y2=16a2 [2002]
  • c)
    x2+y2=4a2
  • d)
    x2+y2=a2
Correct answer is option 'C'. Can you explain this answer?
Verified Answer
The equation of a circle with origin as a centre and passing through e...
Let ABC be an equilateral triangle, whose median is AD.
Given AD = 3a.
In Δ ABD, AB2 = AD2 + BD2 ; ⇒ x2 = 9a2 + (x2/4)  
where AB = BC = AC = x.
In Δ OBD, OB2 = OD2 + BD2
⇒ r2 = 9a2 – 6ar + r2 + 3a2 ;
⇒ 6ar = 12a2 ⇒ r = 2a
So equation of circle is x2 + y2 = 4a2
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Most Upvoted Answer
The equation of a circle with origin as a centre and passing through e...
Let ABC be an equilateral triangle, whose median is AD.
Given AD = 3a.
In Δ ABD, AB2 = AD2 + BD2 ; ⇒ x2 = 9a2 + (x2/4)  
where AB = BC = AC = x.
In Δ OBD, OB2 = OD2 + BD2
⇒ r2 = 9a2 – 6ar + r2 + 3a2 ;
⇒ 6ar = 12a2 ⇒ r = 2a
So equation of circle is x2 + y2 = 4a2
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Community Answer
The equation of a circle with origin as a centre and passing through e...
Equation of the circle passing through equilateral triangle and center at the origin can be written as

X^2+Y^2=r^2

we know centroid divide height in the ratio of 2:1

therefore, radius of the required circle = (2/3)×3a=2a

therefore, equation of the circle ,

X^2+Y^2=4a^2
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