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(C) Let $P = (x, y)$,$A = (0, 2)$,and $B = (-1, 0)$.Given $PA = \frac{1}{\sqrt{2}} PB$,which implies $2 PA^2 = PB^2$.Substituting the coordinates:$2[(

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a circle with centre $(1, 4)$ and radius $\sqrt{10}$ units

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(C) Let $P = (x, y)$,$A = (0, 2)$,and $B = (-1, 0)$.Given $PA = \frac{1}{\sqrt{2}} PB$,which implies $2 PA^2 = PB^2$.Substituting the coordinates:$2[(x - 0)^2 + (y - 2)^2] = (x + 1)^2 + (y - 0)^2$$2(x^2 + y^2 - 4y + 4) = x^2 + 2x + 1 + y^2$$2x^2 + 2y^2 - 8y + 8 = x^2 + y^2 + 2x + 1$$x^2 + y^2 - 2x - 8y + 7 = 0$.This is the equation of a circle in the form $x^2 + y^2 + 2gx + 2fy + c = 0$.Comparing,$2g = -2 \implies g = -1$ and $2f = -8 \implies f = -4$.The centre is $(-g, -f) = (1, 4)$.The radius is $\sqrt{g^2 + f^2 - c} = \sqrt{(-1)^2 + (-4)^2 - 7} = \sqrt{1 + 16 - 7} = \sqrt{10}$ units.

If a point $P$ moves such that the distance from $(0, 2)$ to $P$ is $\frac{1}{\sqrt{2}}$ times the distance of $P$ from $(-1, 0)$,then the locus of the point $P$ is:

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