What will be the circumcentre of a triangle whose sides are 3x - y + 3 = 0, 3x + 4y + 3 = 0 and x + 3y + 11 = 0?

Options

A. (3, 0)
B. (-3, 0)
C. (3, -3
D. (-3, 3)

Answer

To find the circumcenter of the triangle defined by the lines \(3x - y + 3 = 0\), \(3x + 4y + 3 = 0\), and \(x + 3y + 11 = 0\), we will follow these steps:

### Step 1: Find the intersection points of the lines
We need to find the vertices of the triangle formed by the intersection of these lines.

#### Intersection of Line 1 and Line 2:
1. **Equations**:
- Line 1: \(3x - y + 3 = 0\) (or \(y = 3x + 3\))
- Line 2: \(3x + 4y + 3 = 0\) (or \(4y = -3x - 3\) or \(y = -\frac{3}{4}x - \frac{3}{4}\))

2. **Substituting \(y\) from Line 1 into Line 2**:
\[ 3x + 4(3x + 3) + 3 = 0 \]
\[ 3x + 12x + 12 + 3 = 0 \implies 15x + 15 = 0 \implies x = -1 \]
Substituting \(x = -1\) into Line 1:
\[ y = 3(-1) + 3 = 0 \]
**Point A**: \((-1, 0)\)

#### Intersection of Line 2 and Line 3:
1. **Equations**:
- Line 2: \(3x + 4y + 3 = 0\)
- Line 3: \(x + 3y + 11 = 0\) (or \(3y = -x - 11\) or \(y = -\frac{1}{3}x - \frac{11}{3}\))

2. **Substituting \(y\) from Line 3 into Line 2**:
\[ 3x + 4\left(-\frac{1}{3}x - \frac{11}{3}\right) + 3 = 0 \]
\[ 3x - \frac{4}{3}x - \frac{44}{3} + 3 = 0 \]
\[ \left(3 - \frac{4}{3}\right)x + 3 - \frac{44}{3} = 0 \]
\[ \frac{9}{3}x - \frac{4}{3}x + 3 - \frac{44}{3} = 0 \implies \frac{5}{3}x - \frac{35}{3} = 0 \implies x = 7 \]
Substituting \(x = 7\) into Line 3:
\[ y = -\frac{1}{3}(7) - \frac{11}{3} = -\frac{7 + 11}{3} = -6 \]
**Point B**: \((7, -6)\)

#### Intersection of Line 1 and Line 3:
1. **Equations**:
- Line 1: \(3x - y + 3 = 0\)
- Line 3: \(x + 3y + 11 = 0\)

2. **Substituting \(y\) from Line 1 into Line 3**:
\[ x + 3(3x + 3) + 11 = 0 \]
\[ x + 9x + 9 + 11 = 0 \implies 10x + 20 = 0 \implies x = -2 \]
Substituting \(x = -2\) into Line 1:
\[ y = 3(-2) + 3 = -6 \]
**Point C**: \((-2, -3)\)

### Step 2: Calculate the circumcenter
The circumcenter is the point equidistant from all three vertices \(A(-1, 0)\), \(B(7, -6)\), and \(C(-2, -3)\).

1. **Using the distance formula**:
\[ d(A, O) = d(B, O) = d(C, O) \]
Let the circumcenter be \((h, k)\).

2. **Setting up equations**:
- From \(A\) and \(B\):
\[ \sqrt{(h + 1)^2 + (k - 0)^2} = \sqrt{(h - 7)^2 + (k + 6)^2} \]
Squaring both sides:
\[ (h + 1)^2 + k^2 = (h - 7)^2 + (k + 6)^2 \]
Expanding and simplifying gives:
\[ 2h + 1 = -14h + 49 + 36 + 12k \implies 16h + 12k = 84 \quad \text{(Equation 1)} \]

- From \(B\) and \(C\):
\[ \sqrt{(h - 7)^2 + (k + 6)^2} = \sqrt{(h + 2)^2 + (k + 3)^2} \]
Squaring both sides:
\[ (h - 7)^2 + (k + 6)^2 = (h + 2)^2 + (k + 3)^2 \]
Expanding and simplifying gives:
\[ -14h + 49 + 36 + 12k = 4 + 4h + 9 + 6k \implies -18h + 6k = -60 \quad \text{(Equation 2)} \]

### Step 3: Solve the equations
From Equation 1:
\[ 16h + 12k = 84 \implies 4h + 3k = 21 \quad \text{(Equation 3)} \]
From Equation 2:
\[ -18h + 6k = -60 \implies -3h + k = -10 \quad \text{(Equation 4)} \]

Now, we can solve Equations 3 and 4 together:
1. From Equation 4:
\[ k = 3h - 10 \]
2. Substitute \(k\) in Equation 3:
\[ 4h + 3(3h - 10) = 21 \]
\[ 4h + 9h - 30 = 21 \implies 13h = 51 \implies h = 3 \]
3. Substitute \(h = 3\) back into Equation 4:
\[ k = 3(3) - 10 = 9 - 10 = -1 \]

### Final Result
The circumcenter of the triangle is \((3, -1)\).