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(D) Given equations: $x^2+y^2-8x=0$ $(1)$ and $\frac{x^2}{9}-\frac{y^2}{4}=1$ $(2)$.From $(2)$,$4x^2-9y^2=36 \Rightarrow 9y^2=4x^2-36$.Substitute $y^2

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$6x-9y=20$

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(D) Given equations: $x^2+y^2-8x=0$ $(1)$ and $\frac{x^2}{9}-\frac{y^2}{4}=1$ $(2)$.From $(2)$,$4x^2-9y^2=36 \Rightarrow 9y^2=4x^2-36$.Substitute $y^2=8x-x^2$ from $(1)$ into $(2)$: $4x^2-9(8x-x^2)=36$.$4x^2-72x+9x^2=36 \Rightarrow 13x^2-72x-36=0$.$(13x+6)(x-6)=0$,so $x=6$ or $x=-\frac{6}{13}$.For $x=6$,$y^2=8(6)-6^2=48-36=12$,so $y=\pm\sqrt{12}$.Thus,$A=(6, \sqrt{12})$ and $B=(6, -\sqrt{12})$.Let $P=(\alpha, \beta)$ be a point on $2x-3y+4=0$,so $2\alpha-3\beta+4=0$.The centroid $(h, k)$ of $	riangle PAB$ is $h=\frac{6+6+\alpha}{3} = \frac{12+\alpha}{3}$ and $k=\frac{\sqrt{12}-\sqrt{12}+\beta}{3} = \frac{\beta}{3}$.Then $\alpha=3h-12$ and $\beta=3k$.Substitute into $2\alpha-3\beta+4=0$: $2(3h-12)-3(3k)+4=0$.$6h-24-9k+4=0 \Rightarrow 6h-9k=20$.The locus of the centroid is $6x-9y=20$.

If $A$ and $B$ are the points of intersection of the circle $x^2+y^2-8x=0$ and the hyperbola $\frac{x^2}{9}-\frac{y^2}{4}=1$,and a point $P$ moves on the line $2x-3y+4=0$,then the centroid of $\triangle PAB$ lies on the line:

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