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Question Number 141681 by ZiYangLee last updated on 22/May/21

$$\mathrm{On}\mathrm{the}\mathrm{Argand}\mathrm{Diagram},\mathrm{the}\mathrm{variable}\mathrm{point}$$$$\mathrm{Z}\mathrm{represents}\mathrm{a}\mathrm{complex}\mathrm{number}z.$$$$\mathrm{Find}\mathrm{the}\mathrm{equation}\mathrm{of}\mathrm{the}\mathrm{locus}\mathrm{of}\mathrm{a}\mathrm{point}$$$$\mathrm{Z}\mathrm{which}\mathrm{moves}\mathrm{such}\mathrm{that}\mid \frac{z-1}{z+2}\mid =2$$

Answered by MJS_new last updated on 22/May/21

$$\mathrm{generally}$$$$\mathrm{let}z=x+y\mathrm{i}$$$$\mid \frac{x+y\mathrm{i}+a+b\mathrm{i}}{x+y\mathrm{i}+c+d\mathrm{i}}\mid =r;r>0\wedge r\ne 1$$$$\mathrm{leads}\mathrm{to}$$$${(x-\frac{a-{cr}^2}{r^2-1})}^2+{(y-\frac{b-{dr}^2}{r^2-1})}^2=\frac{({(a-c)}^2+{(b-d)}^2)r^2}{{(r^2-1)}^2}$$$$\mathrm{this}\mathrm{is}\mathrm{a}\mathrm{circle}\mathrm{with}\mathrm{center}\left(\begin{array}{c}\frac{a-{cr}^2}{r^2-1}\\ \frac{b-{dr}^2}{r^2-1}\end{array}\right)\mathrm{and}$$$$\mathrm{radius}\frac{r}{r^2-1}\sqrt{{(a-c)}^2+{(b-d)}^2}$$$$\mathrm{for}r=1\mathrm{we}\mathrm{get}\mathrm{a}\mathrm{line}$$$$2(a-c)x+2(b-d)y+a^2+b^2-c^2-d^2=0$$

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