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Question Number 19735 by Tinkutara last updated on 15/Aug/17

$$\mathrm{If}z=\lambda +3+i\sqrt{5-{\lambda }^2},\mathrm{then}\mathrm{the}\mathrm{locus}$$$$\mathrm{of}z\mathrm{is}\mathrm{a}$$

Answered by ajfour last updated on 15/Aug/17

$$\mathrm{if}\mathrm{we}\mathrm{let}\mathrm{z}=\mathrm{x}+\mathrm{iy}$$$$(\mathrm{x}-\lambda -3)+\mathrm{i}(\mathrm{y}-\sqrt{5-{\lambda }^2})=0$$$$\Rightarrow \mathrm{x}=\lambda +3;{\mathrm{y}}^2=5-{\lambda }^2$$$$\mathrm{or}{\mathrm{y}}^2=5-{(\mathrm{x}-3)}^2$$$${(\mathrm{x}-3)}^2+{\mathrm{y}}^2={\left(\sqrt{5}\right)}^2.\left(\mathrm{circle}\right)$$

Commented by Tinkutara last updated on 15/Aug/17

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

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