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Question Number 63517 by Rio Michael last updated on 05/Jul/19

$$\mathrm{Given}\mathrm{that}\mid z-6\mid =2\mid z+6-9i\mid ,$$$$\mathrm{a})\mathrm{Use}\mathrm{algebra}\mathrm{to}\mathrm{show}\mathrm{that}\mathrm{the}\mathrm{locus}\mathrm{of}z\mathrm{is}\mathrm{a}\mathrm{circle},$$$$\mathrm{stating}\mathrm{its}\mathrm{center}\mathrm{and}\mathrm{its}\mathrm{radius}.$$$$\mathrm{b})\mathrm{sketch}\mathrm{the}\mathrm{locus}z\mathrm{on}\mathrm{an}\mathrm{argand}\mathrm{diagram}.$$

Answered by MJS last updated on 05/Jul/19

$$\mid a-6+b\mathrm{i}\mid =2\mid a+6+(b-9)\mathrm{i}\mid$$$$\sqrt{a^2-12a+b^2+36}=2\sqrt{a^2+12a+b^2-18b+117}$$$$a^2-12a+b^2+36=4(a^2+12a+b^2-18b+117)$$$$a^2+20a+b^2-24b+144=0$$$${(a-p)}^2+{(b-q)}^2-r^2=0$$$$a^2-2pa+p^2+b^2-2qb+q^2-r^2=0$$$$-2p=20\Rightarrow p=-10$$$$-2q=-24\Rightarrow q=12$$$$a^2+20a+100+b^2-24b+144-r^2=0$$$$a^2+20a+b^2-24b+144=r^2-100\Rightarrow r^2-100=0\Rightarrow r=10$$$${(a+10)}^2+{(b-12)}^2-{10}^2=0$$$$\Rightarrow \mathrm{center}=\left(\begin{array}{c}-10\\ 12\end{array}\right)\mathrm{radius}=10$$

Commented by Rio Michael last updated on 05/Jul/19

$$correct!$$

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