The circumcentre of a triangle formed by the | vedclass

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(A) The given equations are $xy + 2x + 2y + 4 = 0$ and $x + y + 2 = 0$.Factorizing the first equation: $x(y + 2) + 2(y + 2) = 0 \Rightarrow (x + 2)(y

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$(-1, -1)$

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(A) The given equations are $xy + 2x + 2y + 4 = 0$ and $x + y + 2 = 0$.Factorizing the first equation: $x(y + 2) + 2(y + 2) = 0 \Rightarrow (x + 2)(y + 2) = 0$.This represents two lines: $x = -2$ and $y = -2$.The second equation 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 $(-2, -2)$.$2$. Intersection of $x = -2$ and $x + y + 2 = 0$ is $(-2, 0)$.$3$. Intersection of $y = -2$ and $x + y + 2 = 0$ is $(0, -2)$.The vertices of the triangle are $A(-2, -2)$,$B(-2, 0)$,and $C(0, -2)$.This is a right-angled triangle with the right angle at vertex $A(-2, -2)$ because the lines $x = -2$ and $y = -2$ are perpendicular.In a right-angled triangle,the circumcentre is the midpoint of the hypotenuse $BC$.The hypotenuse connects $B(-2, 0)$ and $C(0, -2)$.Midpoint $= (\frac{-2 + 0}{2}, \frac{0 - 2}{2}) = (-1, -1)$.Thus,the circumcentre is $(-1, -1)$.

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

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