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

Options

A. `x^(2)+y^(2)=9`
B. `x^(2)+y^(2)=16`
C. `x^(2)+y^(2)=4`
D. `x^(2)+y^(2)=1`

Answer

To find the equation of a circle with the origin as the center and passing through the vertices of an equilateral triangle whose altitude is 3 units, 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 altitude of an equilateral triangle can be calculated using the formula:
\[ \text{Altitude} = \frac{\sqrt{3}}{2} \times \text{side} \]
Given that the altitude is 3 units, we can set up the equation to find the side length \( s \).

### Step 2: Calculate the side length of the triangle
Using the altitude formula:
\[ 3 = \frac{\sqrt{3}}{2} \times s \]
Solving for \( s \):
\[ s = \frac{3 \times 2}{\sqrt{3}} = \frac{6}{\sqrt{3}} = 2\sqrt{3} \]

### Step 3: Determine the coordinates of the vertices
The vertices of the equilateral triangle can be positioned in a coordinate system. If we place one vertex at the top, the coordinates of the vertices \( A, B, C \) can be represented as follows:
- Vertex \( A \) (top vertex): \( (0, 3) \)
- Vertex \( B \) (bottom left): \( \left(-\frac{s}{2}, 0\right) = \left(-\sqrt{3}, 0\right) \)
- Vertex \( C \) (bottom right): \( \left(\frac{s}{2}, 0\right) = \left(\sqrt{3}, 0\right) \)

### Step 4: Find the radius of the circle
The radius of the circle is the distance from the origin (0, 0) to any of the vertices. We can calculate the distance to vertex \( A \):
\[ r = \sqrt{(0 - 0)^2 + (3 - 0)^2} = \sqrt{0 + 9} = 3 \]

### Step 5: Write the equation of the circle
The general equation of a circle with center at the origin is given by:
\[ x^2 + y^2 = r^2 \]
Substituting \( r = 3 \):
\[ x^2 + y^2 = 3^2 = 9 \]

### Final Equation
Thus, the equation of the circle is:
\[ \boxed{x^2 + y^2 = 9} \]