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

On the Argand Diagram, the variable point Z represents a complex number z. Find the equation of the locus of a point Z which moves such that ∣((z−1)/(z+2))∣=2

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

Answered by MJS_new last updated on 22/May/21

generally let z=x+yi ∣((x+yi+a+bi)/(x+yi+c+di))∣=r; r>0∧r≠1 leads to (x−((a−cr^2 )/(r^2 −1)))^2 +(y−((b−dr^2 )/(r^2 −1)))^2 =((((a−c)^2 +(b−d)^2 )r^2 )/((r^2 −1)^2 )) this is a circle with center ((((a−cr^2 )/(r^2 −1))),(((b−dr^2 )/(r^2 −1))) ) and radius (r/(r^2 −1))(√((a−c)^2 +(b−d)^2 )) for r=1 we get a line 2(a−c)x+2(b−d)y+a^2 +b^2 −c^2 −d^2 =0

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

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