The circumcentre of the triangle formed by th | vedclass

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(C) The given equation $xy + 2x + 2y + 4 = 0$ can be factored as $(x + 2)(y + 2) = 0$,which represents two lines: $x = -2$ and $y = -2$.The third line

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$(-1, -1)$

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(C) The given equation $xy + 2x + 2y + 4 = 0$ can be factored as $(x + 2)(y + 2) = 0$,which represents two lines: $x = -2$ and $y = -2$.The third line is $x + y + 2 = 0$.To find the vertices of the triangle,we find the intersection points of these lines:$1$. Intersection of $x = -2$ and $y = -2$ is $C(-2, -2)$.$2$. Intersection of $x = -2$ and $x + y + 2 = 0$ is $A(-2, 0)$.$3$. Intersection of $y = -2$ and $x + y + 2 = 0$ is $B(0, -2)$.Since the lines $x = -2$ and $y = -2$ are perpendicular,the triangle is a right-angled triangle with the right angle at $C(-2, -2)$.The circumcentre of a right-angled triangle is the midpoint of the hypotenuse $AB$.The midpoint of $AB$ is $(\frac{-2 + 0}{2}, \frac{0 - 2}{2}) = (-1, -1)$.

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

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