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Question Number 19733 by Tinkutara last updated on 15/Aug/17

$$\mathrm{If}\mid z+1\mid =\sqrt{2}\mid z-1\mid ,\mathrm{then}\mathrm{the}\mathrm{locus}$$$$\mathrm{described}\mathrm{by}\mathrm{the}\mathrm{point}z\mathrm{in}\mathrm{the}\mathrm{argand}$$$$\mathrm{diagram}\mathrm{is}\mathrm{a}$$

Answered by ajfour last updated on 15/Aug/17

$$\mathrm{let}\mathrm{z}=\mathrm{x}+\mathrm{iy}$$$$\mid (\mathrm{x}+1)+\mathrm{iy}{\mid }^2=2\mid (\mathrm{x}-1)+\mathrm{iy}{\mid }^2$$$$\Rightarrow {\mathrm{x}}^2+2\mathrm{x}+1+{\mathrm{y}}^2={2\mathrm{x}}^2-4\mathrm{x}+2+{2\mathrm{y}}^2$$$$\mathrm{or}{\mathrm{x}}^2-6\mathrm{x}+{\mathrm{y}}^2+1=0$$$${(\mathrm{x}-3)}^2+{\mathrm{y}}^2={\left(2\sqrt{2}\right)}^2$$$$\mathrm{or}\mid \mathrm{z}-3\mid =2\sqrt{2}.$$

Commented by Tinkutara last updated on 15/Aug/17

$$\mathrm{Thank}\mathrm{you}\mathrm{very}\mathrm{much}\mathrm{Sir}!$$

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