Origin is the centre of a circle passing through the vertices of an equilateral triangie whose median is of length 3a then the equation of the circle is

Options

A. `x^(2)+y^(2)=a^(2)`
B. `x^(2)+y^(2)=2a^(2)`
C. `x^(2)+y^(2)=3a^(2)`
D. `x^(2)+y^(2)=4a^(2)`

Answer

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

### Step 1: Understand the properties of the triangle
An equilateral triangle has all sides equal and all angles equal to \(60^\circ\). The median of an equilateral triangle also acts as the altitude and bisects the triangle into two \(30-60-90\) triangles.

### Step 2: Use the median length to find the side length
The length of the median \(m\) of an equilateral triangle with side length \(s\) is given by the formula:
\[ m = \frac{\sqrt{3}}{2} s \]
Given that the median length is \(3a\), we can set up the equation:
\[ \frac{\sqrt{3}}{2} s = 3a \]
From this, we can solve for \(s\):
\[ s = \frac{2 \cdot 3a}{\sqrt{3}} = \frac{6a}{\sqrt{3}} = 2\sqrt{3}a \]

### Step 3: Determine the circumradius of the triangle
The circumradius \(R\) of an equilateral triangle is given by the formula:
\[ R = \frac{s}{\sqrt{3}} \]
Substituting the value of \(s\):
\[ R = \frac{2\sqrt{3}a}{\sqrt{3}} = 2a \]

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

### Conclusion
Thus, the equation of the circle is:
\[ x^2 + y^2 = 4a^2 \]