Show that the equation of the circle with centre at origin and passing through the vertices of an equilateral triangle whose median is of length 3a is `x^(2)+y^(2)=4a^(2)`.

Answer

To solve the problem, we need to show that the equation of the circle with its center at the origin and passing through the vertices of an equilateral triangle, whose median is of length \(3a\), is given by the equation \(x^2 + y^2 = 4a^2\).

### Step-by-Step Solution:

1. **Understanding the Median of an Equilateral Triangle**:
The median of an equilateral triangle divides it into two equal halves. The length of the median \(m\) in terms of the side length \(s\) of the triangle is given by the formula:
\[ m = \frac{\sqrt{3}}{2} s \]
Given that the median is \(3a\), we can set this equal to the formula:
\[ \frac{\sqrt{3}}{2} s = 3a \]

2. **Finding the Side Length of the Triangle**:
Rearranging the equation to solve for \(s\):
\[ s = \frac{2 \cdot 3a}{\sqrt{3}} = \frac{6a}{\sqrt{3}} = 2\sqrt{3}a \]

3. **Finding the Circumradius of the Triangle**:
The circumradius \(R\) of an equilateral triangle can be calculated using the formula:
\[ R = \frac{s}{\sqrt{3}} \]
Substituting the value of \(s\):
\[ R = \frac{2\sqrt{3}a}{\sqrt{3}} = 2a \]

4. **Equation of the Circle**:
The equation of a circle with center at the origin and radius \(R\) is given by:
\[ x^2 + y^2 = R^2 \]
Substituting \(R = 2a\):
\[ x^2 + y^2 = (2a)^2 = 4a^2 \]

5. **Conclusion**:
Thus, we have shown that the equation of the circle with its center at the origin and passing through the vertices of the equilateral triangle is:
\[ x^2 + y^2 = 4a^2 \]