The equation of the circumcircle of the triangle formed by the lines `y+sqrt(3)x=6,y-sqrt(3)` x=6 and y=0 is

Options

A. `x^(2)+y^(2)-4y=0`
B. `x^(2)+y^(2)+4x=0`
C. `x^(2)+y^(2)-4y-12=0`
D. `x^(2)+y^(2)+4x=12`

Answer

To find the equation of the circumcircle of the triangle formed by the lines \( y + \sqrt{3}x = 6 \), \( y - \sqrt{3}x = 6 \), and \( y = 0 \), we will follow these steps:

### Step 1: Find the points of intersection of the lines
1. **Intersection of \( y + \sqrt{3}x = 6 \) and \( y = 0 \)**:
\[ 0 + \sqrt{3}x = 6 \implies x = \frac{6}{\sqrt{3}} = 2\sqrt{3} \]
So, the point is \( A(2\sqrt{3}, 0) \).

2. **Intersection of \( y - \sqrt{3}x = 6 \) and \( y = 0 \)**:
\[ 0 - \sqrt{3}x = 6 \implies x = -\frac{6}{\sqrt{3}} = -2\sqrt{3} \]
So, the point is \( B(-2\sqrt{3}, 0) \).

3. **Intersection of \( y + \sqrt{3}x = 6 \) and \( y - \sqrt{3}x = 6 \)**:
Set both equations equal to each other:
\[ y + \sqrt{3}x = y - \sqrt{3}x \implies 2\sqrt{3}x = 0 \implies x = 0 \]
Substitute \( x = 0 \) into either equation:
\[ y = 6 \]
So, the point is \( C(0, 6) \).

### Step 2: Identify the vertices of the triangle
The vertices of the triangle are:
- \( A(2\sqrt{3}, 0) \)
- \( B(-2\sqrt{3}, 0) \)
- \( C(0, 6) \)

### Step 3: Find the circumcenter
The circumcenter of an equilateral triangle is the same as its centroid. The coordinates of the centroid \( G \) can be calculated as:
\[ G\left(\frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3}\right) \]
Substituting the coordinates:
\[ G\left(\frac{2\sqrt{3} + (-2\sqrt{3}) + 0}{3}, \frac{0 + 0 + 6}{3}\right) = G(0, 2) \]

### Step 4: Calculate the circumradius
The circumradius \( R \) of an equilateral triangle can be calculated using the formula:
\[ R = \frac{a}{\sqrt{3}} \]
where \( a \) is the length of a side. First, we need to find the length of side \( AB \):
\[ AB = \sqrt{(2\sqrt{3} - (-2\sqrt{3}))^2 + (0 - 0)^2} = \sqrt{(4\sqrt{3})^2} = 12 \]
Thus, the circumradius is:
\[ R = \frac{12}{\sqrt{3}} = 4\sqrt{3} \]

### Step 5: Write the equation of the circumcircle
The equation of a circle with center \( (h, k) \) and radius \( r \) is given by:
\[ (x - h)^2 + (y - k)^2 = r^2 \]
Substituting \( h = 0 \), \( k = 2 \), and \( r = 4\sqrt{3} \):
\[ (x - 0)^2 + (y - 2)^2 = (4\sqrt{3})^2 \]
This simplifies to:
\[ x^2 + (y - 2)^2 = 48 \]
Expanding this gives:
\[ x^2 + y^2 - 4y + 4 = 48 \]
Rearranging leads to:
\[ x^2 + y^2 - 4y - 44 = 0 \]

### Final Answer
The equation of the circumcircle is:
\[ x^2 + y^2 - 4y - 44 = 0 \]