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 \( y = 0 \)**:
- Substitute \( y = 0 \) into the first equation:
\[ 0 + \sqrt{3}x = 6 \implies \sqrt{3}x = 6 \implies x = \frac{6}{\sqrt{3}} = 2\sqrt{3} \]
- So, the point is \( A(2\sqrt{3}, 0) \).

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

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

### Step 2: Find the circumcenter
The circumcenter of a triangle can be found as the average of the vertices:
- Coordinates of points \( A(2\sqrt{3}, 0) \), \( B(-2\sqrt{3}, 0) \), and \( C(0, 6) \).

\[ \text{Circumcenter} (O) = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) = \left( \frac{2\sqrt{3} - 2\sqrt{3} + 0}{3}, \frac{0 + 0 + 6}{3} \right) = \left( 0, 2 \right) \]

### Step 3: Find the circumradius
To find the circumradius \( R \), we can use the formula:
\[ R = \frac{a}{2\sin A} \]
where \( a \) is the length of a side of the triangle and \( A \) is the angle opposite that side.

1. **Calculate the length of side \( AB \)**:
\[ AB = \sqrt{(2\sqrt{3} - (-2\sqrt{3}))^2 + (0 - 0)^2} = \sqrt{(4\sqrt{3})^2} = 4\sqrt{3} \]

2. **Angle \( A \) in an equilateral triangle**:
Since the triangle is symmetric about the y-axis, it is an isosceles triangle. The angles at points \( A \) and \( B \) are \( 60^\circ \).

3. **Calculate \( R \)**:
\[ R = \frac{4\sqrt{3}}{2 \cdot \frac{\sqrt{3}}{2}} = \frac{4\sqrt{3}}{\sqrt{3}} = 4 \]

### Step 4: 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 \( (h, k) = (0, 2) \) and \( r = 4 \):
\[ (x - 0)^2 + (y - 2)^2 = 4^2 \implies x^2 + (y - 2)^2 = 16 \]
Expanding this:
\[ x^2 + (y^2 - 4y + 4) = 16 \implies x^2 + y^2 - 4y - 12 = 0 \]

### Final Answer:
The equation of the circumcircle is:
\[ x^2 + y^2 - 4y - 12 = 0 \]