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(C) Let $P = (x, y)$. The given condition is $PA = 2PB$.$\sqrt{(x-4)^2 + y^2} = 2\sqrt{(x+4)^2 + y^2}$.Squaring both sides: $(x-4)^2 + y^2 = 4((x+4)^2

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$\frac{32}{3}$

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(C) Let $P = (x, y)$. The given condition is $PA = 2PB$.$\sqrt{(x-4)^2 + y^2} = 2\sqrt{(x+4)^2 + y^2}$.Squaring both sides: $(x-4)^2 + y^2 = 4((x+4)^2 + y^2)$.$x^2 - 8x + 16 + y^2 = 4(x^2 + 8x + 16 + y^2)$.$x^2 - 8x + 16 + y^2 = 4x^2 + 32x + 64 + 4y^2$.$3x^2 + 3y^2 + 40x + 48 = 0$.Dividing by $3$: $x^2 + y^2 + \frac{40}{3}x + 16 = 0$.This is a circle with center $O' = (-\frac{20}{3}, 0)$ and radius $r = \sqrt{(-\frac{20}{3})^2 - 16} = \sqrt{\frac{400}{9} - 16} = \sqrt{\frac{400-144}{9}} = \sqrt{\frac{256}{9}} = \frac{16}{3}$.The given line is $3y - 3x - 20 = 0$,which can be written as $y - x - \frac{20}{3} = 0$.Check if the center $(-\frac{20}{3}, 0)$ lies on the line: $0 - (-\frac{20}{3}) - \frac{20}{3} = 0$. Yes,it does.Since the line passes through the center of the circle,the chord $CD$ is a diameter.Distance $CD = 2r = 2 	imes \frac{16}{3} = \frac{32}{3}$.

$P$ is a variable point such that the distance of $P$ from $A(4,0)$ is twice the distance of $P$ from $B(-4,0)$. If the line $3y - 3x - 20 = 0$ intersects the locus of $P$ at the points $C$ and $D$,then the distance between $C$ and $D$ is:

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