Question

Equation of the circle with origin as centre and passing through the vertices of an equilateral triangle whose median is of length \(3\) is:

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

Hint:

Important formulas for an equilateral triangle of side \(s\): \[ \text{Median}=\frac{\sqrt3}{2}s \] \[ \text{Circumradius}=\frac{s}{\sqrt3} \] Circle centered at origin with radius \(R\): \[ x^2+y^2=R^2 \]

Answer 1

The Correct Option is A

Concept: In an equilateral triangle: \[ \text{Median}=\text{Altitude}=\frac{\sqrt3}{2}s \] where \(s\) is the side length. Also, the centroid, circumcenter and incentre coincide at the same point. The circumradius of an equilateral triangle is: \[ R=\frac{s}{\sqrt3} \] Since the circle is centered at the origin and passes through all vertices, the radius of the circle is the circumradius.

Step 1: Use the median formula. Given: \[ \text{Median}=3a \] Now: \[ \frac{\sqrt3}{2}s=3a \] Solving for \(s\): \[ s=\frac{6a}{\sqrt3} \] \[ s=2\sqrt3\,a \]

Step 2: Find the circumradius. \[ R=\frac{s}{\sqrt3} \] Substituting: \[ R=\frac{2\sqrt3\,a}{\sqrt3} \] \[ R=2a \]

Step 3: Write the equation of the circle. Circle centered at origin with radius \(R\) has equation: \[ x^2+y^2=R^2 \] Substituting \(R=2a\): \[ x^2+y^2=(2a)^2 \] \[ x^2+y^2=4a^2 \] Hence the required equation is: \[ \boxed{x^2+y^2=4a^2} \]