From a point A(1, 0) on the circle x^2+y^2-2x | vedclass

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(A) Let the point $P(h, k)$,$A(1, 0)$,and $B(x_1, y_1)$. We have $AP=3AB$. Since $P$ lies on the extension of $AB$,$B$ lies between $A$ and $P$. Thus,

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$x^2+y^2-2x+6y+1=0$

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(A) Let the point $P(h, k)$,$A(1, 0)$,and $B(x_1, y_1)$. We have $AP=3AB$. Since $P$ lies on the extension of $AB$,$B$ lies between $A$ and $P$. Thus,$AP = AB + BP = 3AB$,which implies $BP = 2AB$. Therefore,$B$ divides $AP$ in the ratio $1:2$ internally. Using the section formula,the coordinates of $B(x_1, y_1)$ are: $x_1 = \frac{1 \cdot h + 2 \cdot 1}{1+2} = \frac{h+2}{3}$ $y_1 = \frac{1 \cdot k + 2 \cdot 0}{1+2} = \frac{k}{3}$ Since $B(x_1, y_1)$ lies on the circle $x^2+y^2-2x+2y+1=0$,we substitute these values: $(\frac{h+2}{3})^2 + (\frac{k}{3})^2 - 2(\frac{h+2}{3}) + 2(\frac{k}{3}) + 1 = 0$ Multiply by $9$: $(h+2)^2 + k^2 - 6(h+2) + 6k + 9 = 0$ $h^2 + 4h + 4 + k^2 - 6h - 12 + 6k + 9 = 0$ $h^2 + k^2 - 2h + 6k + 1 = 0$ Replacing $(h, k)$ with $(x, y)$,the locus of $P$ is $x^2+y^2-2x+6y+1=0$.

From a point $A(1, 0)$ on the circle $x^2+y^2-2x+2y+1=0$,a chord $AB$ is drawn and it is extended to a point $P$ such that $AP=3AB$. The equation of the locus of $P$ is

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