Find the equation of a circle with origin as centre and which circumscribes equilateral triangle whose median of length `3a`

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 the origin as the center that circumscribes an equilateral triangle with a median of length \(3a\), we can follow these steps:

### Step 1: Understand the properties of the equilateral triangle
In an equilateral triangle, the centroid (G) divides each median into a ratio of 2:1. This means that if the length of the median is \(3a\), then the segment from a vertex to the centroid (AG) is \( \frac{2}{3} \) of the median.

### Step 2: Calculate the length of AG
Given that the length of the median (AD) is \(3a\):
\[ AG = \frac{2}{3} \times \text{(length of median)} = \frac{2}{3} \times 3a = 2a \]

### Step 3: Determine the radius of the circumcircle
For an equilateral triangle, the radius (R) of the circumcircle is equal to the length from the centroid to a vertex (AG). Thus, the radius of the circumcircle is:
\[ R = AG = 2a \]

### Step 4: Write the equation of the circle
The general equation of a circle with center at the origin (0, 0) and radius \(R\) is given by:
\[ x^2 + y^2 = R^2 \]
Substituting the value of \(R\):
\[ x^2 + y^2 = (2a)^2 = 4a^2 \]

### Final Answer
The equation of the circle is:
\[ \boxed{x^2 + y^2 = 4a^2} \]

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