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(B) Let the vertices of the square be $P, Q, R, S$. The sides are $x=-5$ and $y=4$. The center of the square is $C(3,-4)$.The distance from the center

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$(-13,-4)$

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(B) Let the vertices of the square be $P, Q, R, S$. The sides are $x=-5$ and $y=4$. The center of the square is $C(3,-4)$.The distance from the center $C(3,-4)$ to the line $x=-5$ is $|3 - (-5)| = 8$. Since the side length is $2 	imes 8 = 16$,the other sides are $x=11$ and $y=-12$.The vertices are $P(-5, 4)$,$S(11, 4)$,$R(11, -12)$,and $Q(-5, -12)$.The vertices lying on $x=-5$ are $P(-5, 4)$ and $Q(-5, -12)$.The circumcircle of the square has center $C(3,-4)$ and radius $r = \sqrt{(3 - (-5))^2 + (-4 - 4)^2} = \sqrt{8^2 + (-8)^2} = \sqrt{64 + 64} = \sqrt{128} = 8\sqrt{2}$.The equation of the circle is $(x-3)^2 + (y+4)^2 = 128$.The tangent at $P(-5, 4)$ is $(x-3)(-5-3) + (y+4)(4+4) = 128$ $\Rightarrow -8(x-3) + 8(y+4) = 128$ $\Rightarrow -x+3+y+4 = 16$ $\Rightarrow y-x = 9$.The tangent at $Q(-5, -12)$ is $(x-3)(-5-3) + (y+4)(-12+4) = 128$ $\Rightarrow -8(x-3) - 8(y+4) = 128$ $\Rightarrow -x+3-y-4 = 16$ $\Rightarrow -x-y = 17$ $\Rightarrow x+y = -17$.Solving $y-x=9$ and $x+y=-17$: Adding them gives $2y = -8 \Rightarrow y = -4$. Substituting $y=-4$ into $y-x=9$ gives $-4-x=9 \Rightarrow x = -13$.The point of intersection is $(-13, -4)$.

Two sides of a square are along the lines $x=-5$ and $y=4$. The point of intersection of the diagonals is $(3,-4)$. The point of intersection of the tangents drawn to the circumcircle of the square at the two consecutive vertices lying on $x=-5$ is

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