-1, -1)
b)(1, 1)
c)(-1, 1)
d)(1, -1)

To find the circumcentre of a triangle formed by three lines, we need to find the intersection point of the perpendicular bisectors of any two sides of the triangle.

First, let's find the equation of the line perpendicular to xy + 2x + 2y + 4 = 0. The slope of this line would be the negative reciprocal of the slope of xy + 2x + 2y + 4 = 0.

The slope of xy + 2x + 2y + 4 = 0 can be found by rearranging the equation to slope-intercept form (y = mx + b), where m is the slope:
xy + 2x + 2y + 4 = 0
xy + 2y = -2x - 4
y(x + 2) = -2(x + 2)
y = -2(x + 2)/(x + 2)
y = -2

The slope of xy + 2x + 2y + 4 = 0 is -2, so the slope of the line perpendicular to it is 1/2.

Now, let's find the equation of the line perpendicular to x + y + 2 = 0. The slope of this line would be the negative reciprocal of the slope of x + y + 2 = 0.

The slope of x + y + 2 = 0 is -1, so the slope of the line perpendicular to it is 1.

Now we have two lines:
Line 1: y = (1/2)x + b1
Line 2: y = x + b2

To find the intersection point of these two lines, we can set them equal to each other:
(1/2)x + b1 = x + b2
(1/2)x - x = b2 - b1
(-1/2)x = b2 - b1
x = 2(b1 - b2)

Substituting this value of x into either of the equations, we can solve for y:
y = (1/2)(2(b1 - b2)) + b1
y = b1 - b2 + b1
y = 2b1 - b2

Since the circumcentre is the intersection point of the perpendicular bisectors, we need to find the midpoint of the line segment between two points (x1, y1) and (x2, y2), where (x1, y1) and (x2, y2) are points on the two lines.

The midpoint formula is:
(x_mid, y_mid) = ((x1 + x2)/2, (y1 + y2)/2)

Let's take two points on Line 1, (0, b1) and (-4, 2b1 - 4):
(x_mid, y_mid) = ((0 + -4)/2, (b1 + (2b1 - 4))/2)
(x_mid, y_mid) = (-2, (3b1 - 4)/2)
(x_mid, y_mid) = (-2, (3b1 - 4)/2)

Let's take two points on Line 2, (-2