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(B) Let the general equation of the circle be $x^2 + y^2 + 2gx + 2fy + c = 0$.Since the circle passes through $(2, -2)$,$(-1, -1)$,and $(5, 2)$,we hav

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$x^2 + y^2 - 3x - 3y - 8 = 0$

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(B) Let the general equation of the circle be $x^2 + y^2 + 2gx + 2fy + c = 0$.Since the circle passes through $(2, -2)$,$(-1, -1)$,and $(5, 2)$,we have:$1) \; 4 + 4 + 4g - 4f + c = 0 \Rightarrow 4g - 4f + c = -8$$2) \; 1 + 1 - 2g - 2f + c = 0 \Rightarrow -2g - 2f + c = -2$$3) \; 25 + 4 + 10g + 4f + c = 0 \Rightarrow 10g + 4f + c = -29$Subtracting $(2)$ from $(1)$: $6g - 2f = -6$ $\Rightarrow 3g - f = -3$ $\Rightarrow f = 3g + 3$.Subtracting $(2)$ from $(3)$: $12g + 6f = -27 \Rightarrow 4g + 2f = -9$.Substituting $f = 3g + 3$ into $4g + 2f = -9$: $4g + 2(3g + 3) = -9$ $\Rightarrow 10g + 6 = -9$ $\Rightarrow 10g = -15$ $\Rightarrow g = -1.5$.Then $f = 3(-1.5) + 3 = -4.5 + 3 = -1.5$.Substituting $g$ and $f$ into $(2)$: $-2(-1.5) - 2(-1.5) + c = -2$ $\Rightarrow 3 + 3 + c = -2$ $\Rightarrow c = -8$.The equation is $x^2 + y^2 - 3x - 3y - 8 = 0$.

If the vertices of a triangle are $(2, -2)$,$(-1, -1)$,and $(5, 2)$,then the equation of its circumcircle is

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