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 the lines $x=-2$ and $y=-2$. Given the third line is $x+y+2=0

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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 the lines $x=-2$ and $y=-2$. Given the third line is $x+y+2=0$. To find the vertices of the triangle,we solve the equations in pairs: $1$. Intersection of $x=-2$ and $y=-2$ gives vertex $C(-2,-2)$. $2$. Intersection of $x=-2$ and $x+y+2=0$ gives $-2+y+2=0$,so $y=0$. Vertex $A(-2,0)$. $3$. Intersection of $y=-2$ and $x+y+2=0$ gives $x-2+2=0$,so $x=0$. Vertex $B(0,-2)$. The vertices are $A(-2,0)$,$B(0,-2)$,and $C(-2,-2)$. Since the lines $x=-2$ and $y=-2$ are perpendicular,$\Delta ABC$ 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 $AB$. Midpoint of $AB = (\frac{-2+0}{2}, \frac{0-2}{2}) = (-1,-1)$.

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

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