The circumradius of the triangle formed by th | vedclass

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(C) First,factor the equation $xy+2x+2y+4=0$. $x(y+2)+2(y+2)=0$ $(x+2)(y+2)=0$. This represents two lines: $x=-2$ and $y=-2$. The third line is given

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$\sqrt{2}$ units

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(C) First,factor the equation $xy+2x+2y+4=0$. $x(y+2)+2(y+2)=0$ $(x+2)(y+2)=0$. This represents two lines: $x=-2$ and $y=-2$. The third line is given as $x+y+2=0$,which can be written as $y=-x-2$. To find the vertices of the triangle,we find the intersection points of these lines: $1$. Intersection of $x=-2$ and $y=-2$ is $A(-2, -2)$. $2$. Intersection of $x=-2$ and $y=-x-2$ is $B(-2, 0)$. $3$. Intersection of $y=-2$ and $y=-x-2$ is $C(0, -2)$. The triangle is a right-angled triangle with the right angle at $A(-2, -2)$. The hypotenuse is the segment $BC$ connecting $(-2, 0)$ and $(0, -2)$. The length of the hypotenuse $BC = \sqrt{(0 - (-2))^2 + (-2 - 0)^2} = \sqrt{2^2 + (-2)^2} = \sqrt{4+4} = \sqrt{8} = 2\sqrt{2}$. The circumradius of a right-angled triangle is half the length of the hypotenuse. Circumradius $R = \frac{BC}{2} = \frac{2\sqrt{2}}{2} = \sqrt{2}$ units.

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

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