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

Options

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

Answer

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

### Step 1: Understand the properties of the equilateral triangle
An equilateral triangle has all sides equal and all angles equal to 60 degrees. The median of an equilateral triangle also serves as the altitude and the angle bisector.

### Step 2: Relate the median to 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 is 3a, we can set up the equation:
\[ \frac{\sqrt{3}}{2} s = 3a \]

### Step 3: Solve for the side length (s)
Rearranging the equation to find s:
\[ s = \frac{3a \cdot 2}{\sqrt{3}} = \frac{6a}{\sqrt{3}} = 2\sqrt{3}a \]

### Step 4: Find the circumradius (R)
The circumradius (R) of an equilateral triangle is related to the side length (s) by the formula:
\[ R = \frac{s}{\sqrt{3}} \]
Substituting the value of s:
\[ R = \frac{2\sqrt{3}a}{\sqrt{3}} = 2a \]

### Step 5: 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:
\[ x^2 + y^2 = (2a)^2 = 4a^2 \]

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

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