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 that passes through the vertices of an equilateral triangle with an altitude of 3 units, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Properties of the Equilateral Triangle**:
- The altitude of an equilateral triangle divides it into two 30-60-90 right triangles. The length of the altitude (h) is given as 3 units.

2. **Finding the Side Length of the Triangle**:
- For an equilateral triangle, the relationship between the side length (s) and the altitude (h) is given by:
\[ h = \frac{\sqrt{3}}{2} s \]
- Substituting the given altitude:
\[ 3 = \frac{\sqrt{3}}{2} s \]
- Solving for s:
\[ s = \frac{3 \cdot 2}{\sqrt{3}} = 2\sqrt{3} \]

3. **Finding the Radius of the Circumcircle**:
- The radius (R) of the circumcircle of an equilateral triangle is given by:
\[ R = \frac{s}{\sqrt{3}} \]
- Substituting the side length:
\[ R = \frac{2\sqrt{3}}{\sqrt{3}} = 2 \]

4. **Equation of the Circle**:
- The general equation of a circle with the center at the origin (0,0) and radius R is:
\[ x^2 + y^2 = R^2 \]
- Substituting R = 2:
\[ x^2 + y^2 = 2^2 = 4 \]

5. **Final Answer**:
- The equation of the circle is:
\[ x^2 + y^2 = 4 \]