The equation of a circle with origin as centre and passing through the vertices of an equilateral triangle whose median is of length 3 a is

Options

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)`

Explanation

To find the equation of a circle with its center at the origin and passing through the vertices of an equilateral triangle whose median is of length \(3a\), we can follow these steps:

### Step-by-Step Solution

1. **Understand the Problem**: We need to find the equation of a circle centered at the origin (0, 0) that passes through the vertices of an equilateral triangle. The median of the triangle is given as \(3a\).

2. **Find the Length of the Radius**:
- The centroid of an equilateral triangle divides each median into a ratio of 2:1. Therefore, the distance from the centroid to a vertex (which is the radius of the circumcircle) is \(\frac{2}{3}\) of the length of the median.
- Given that the length of the median is \(3a\), we can calculate the radius \(r\) as follows:
\[ r = \frac{2}{3} \times \text{median} = \frac{2}{3} \times 3a = 2a. \]

3. **Write the Equation of the Circle**:
- The standard equation of a circle with center at the origin (0, 0) and radius \(r\) is given by:
\[ x^2 + y^2 = r^2. \]
- Substituting \(r = 2a\) into the equation:
\[ x^2 + y^2 = (2a)^2 = 4a^2. \]

4. **Final Equation**:
- Therefore, the equation of the circle is:
\[ x^2 + y^2 = 4a^2. \]

### Conclusion
The equation of the circle with the origin as the center and passing through the vertices of the equilateral triangle is:
\[ \boxed{x^2 + y^2 = 4a^2}. \]