The circumcentre of the triangle formed by the lines `xy+2x+2y+4=0` and `x+y+2=0` is

Options

A. (0,0)
B. (-1,-1)
C. (-1,-2)
D. (-2,-2)

Answer

To find the circumcentre of the triangle formed by the lines \(xy + 2x + 2y + 4 = 0\) and \(x + y + 2 = 0\), we can follow these steps:

### Step 1: Rewrite the first equation
The first equation is \(xy + 2x + 2y + 4 = 0\). We can factor this equation:
\[ xy + 2x + 2y + 4 = (x + 2)(y + 2) = 0 \]
This gives us two lines:
1. \(x + 2 = 0\) or \(x = -2\)
2. \(y + 2 = 0\) or \(y = -2\)

### Step 2: Identify the second line
The second line is given as:
\[ x + y + 2 = 0 \]
We can rearrange this to find \(y\):
\[ y = -x - 2 \]

### Step 3: Find the intersection points
Now we need to find the intersection points of the lines:
1. \(x = -2\) and \(y = -2\)
2. \(x = -2\) and \(y = -x - 2\)
3. \(y = -2\) and \(y = -x - 2\)

**Intersection of \(x = -2\) and \(y = -2\):**
\[ (-2, -2) \]

**Intersection of \(x = -2\) and \(y = -x - 2\):**
Substituting \(x = -2\):
\[ y = -(-2) - 2 = 0 \quad \Rightarrow \quad (-2, 0) \]

**Intersection of \(y = -2\) and \(y = -x - 2\):**
Substituting \(y = -2\):
\[ -2 = -x - 2 \quad \Rightarrow \quad x = 0 \quad \Rightarrow \quad (0, -2) \]

### Step 4: Identify the vertices of the triangle
The vertices of the triangle formed by the lines are:
1. \((-2, -2)\)
2. \((-2, 0)\)
3. \((0, -2)\)

### Step 5: Calculate the circumcentre
The circumcentre of a right triangle is the midpoint of the hypotenuse. The hypotenuse is between the points \((-2, 0)\) and \((0, -2)\).

To find the midpoint:
\[ \text{Midpoint} = \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right) = \left(\frac{-2 + 0}{2}, \frac{0 - 2}{2}\right) = \left(-1, -1\right) \]

### Final Answer
The circumcentre of the triangle is:
\[ \boxed{(-1, -1)} \]