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(D) The given circle is $x^2+y^2-4x-4y-8=0$. The center $C$ is $(2,2)$ and the radius $r$ is $\sqrt{2^2+2^2-(-8)} = \sqrt{4+4+8} = 4$. Let $M$ be the

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$2y+3=0$

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(D) The given circle is $x^2+y^2-4x-4y-8=0$. The center $C$ is $(2,2)$ and the radius $r$ is $\sqrt{2^2+2^2-(-8)} = \sqrt{4+4+8} = 4$. Let $M$ be the midpoint of the chord $AB$. Since $PA=PB=2$,$P$ lies on the perpendicular bisector of $AB$. In $	riangle PAM$,$AM = \sqrt{PA^2 - PM^2}$. Since $PA=2$,$AM = \sqrt{4-PM^2}$. Also,in the circle,$AM = \sqrt{r^2 - CM^2} = \sqrt{16 - CM^2}$. Equating the two,$4-PM^2 = 16-CM^2$,so $CM^2 - PM^2 = 12$. Let the line $AB$ have the equation $a(x-2) + b(y+2) = 0$. Since $P(2,-2)$ is on the line,the distance from $C(2,2)$ to the line is $CM = \frac{|a(2-2) + b(2+2)|}{\sqrt{a^2+b^2}} = \frac{4|b|}{\sqrt{a^2+b^2}}$. The distance $PM$ is the distance from $P(2,-2)$ to the line,which is $0$ as $P$ lies on the line. Thus $CM^2 = 12$,so $\frac{16b^2}{a^2+b^2} = 12$,which simplifies to $16b^2 = 12a^2 + 12b^2$,or $4b^2 = 12a^2$,so $b^2 = 3a^2$,$b = \pm \sqrt{3}a$. However,checking the options,the line $2x+3=0$ is a vertical line $x=-1.5$. The distance from $C(2,2)$ to $x=-1.5$ is $3.5$. $CM^2 = 12.25$. $PM^2$ is the distance from $P(2,-2)$ to $x=-1.5$,which is $3.5^2 = 12.25$. $CM^2-PM^2=0 
eq 12$. Re-evaluating: The line $AB$ must be perpendicular to the radius $CP$. The slope of $CP$ is $\frac{2-(-2)}{2-2} = \infty$. Thus $AB$ is a horizontal line $y=k$. Since $P(2,-2)$ is on the circle,and $PA=PB=2$,$P$ is the midpoint of the arc $AB$. The line $AB$ is $y=-2+h$. Testing $2y+3=0 \implies y=-1.5$. Distance from $C(2,2)$ to $y=-1.5$ is $3.5$. $CM^2 = 12.25$. Distance from $P(2,-2)$ to $y=-1.5$ is $0.5$. $PM^2 = 0.25$. $CM^2-PM^2 = 12.25-0.25 = 12$. This matches.

$A$ line meets the circle $x^2+y^2-4x-4y-8=0$ at two points $A$ and $B$. If $P(2,-2)$ is a point on the circle such that $PA=PB=2$,then the equation of the line $AB$ is:

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