Question

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

• A. \(x^{2}+y^{2}=9a^{2}\)
• B. \(x^{2}+y^{2}=16a^{2}\)
• C. \(x^{2}+y^{2}=4a^{2}\)
• D. \(x^{2}+y^{2}=a^{2}\)

Hint:

The centre of the circumcircle of an equilateral triangle is its centroid, which divides each median in the ratio 2:1.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept
For an equilateral triangle the centroid, circumcentre and orthocentre coincide.

Step 2: Radius
The centroid divides the median in the ratio \(2:1\), so the circumradius is \(\dfrac{2}{3}\) of the median.
\(R=\dfrac{2}{3}\times 3a=2a\).

Step 3: Equation
The centre is the origin, so \(x^{2}+y^{2}=R^{2}=4a^{2}\).

Final Answer:
The circle is x^2 + y^2 = 4a^2. \[ \boxed{\text{(C) }x^{2}+y^{2}=4a^{2}} \]