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Question Number 109305 by peter frank last updated on 22/Aug/20

Answered by Aziztisffola last updated on 22/Aug/20

$$\mathrm{let}\mathrm{A}(4;0)\mathrm{and}\mathrm{P}(x;y)\in \mathcal{C}(\mathrm{o};2)$$$$\Rightarrow x^2+y^2=4\Rightarrow y=\underset{-}{+}\sqrt{4-x^2}$$$$\mathrm{let}\mathrm{I}\mathrm{be}\mathrm{center}\mathrm{of}\left[\mathrm{AP}\right]$$$$\mathrm{I}(\frac{x+4}{2};\frac{\sqrt{4-x^2}}{2})\wedge \mathrm{I}(\frac{x+4}{2};-\frac{\sqrt{4-x^2}}{2})$$

Commented by Aziztisffola last updated on 22/Aug/20

Answered by 1549442205PVT last updated on 22/Aug/20

$$\mathrm{P}({\mathrm{x}}_1,{\mathrm{y}}_1)\in (\mathrm{O},2)\Rightarrow {\mathrm{x}}_1^2+{\mathrm{y}}_1^2=4\Rightarrow {\mathrm{y}}_1=\pm \sqrt{4-{\mathrm{x}}^2}$$$$\Rightarrow \mathrm{P}({\mathrm{x}}_1,\pm \sqrt{4-{\mathrm{x}}_1^2}),\mathrm{I}-\mathrm{midpoint}\mathrm{of}\mathrm{AP}$$$$(\mathrm{since}\mathrm{by}\mathrm{the}\mathrm{hypothesis}\mathrm{AI}:\mathrm{AP}=1:2)$$$$\mathrm{since}\mathrm{A}(4,0),\mathrm{we}\mathrm{have}\mathrm{I}(\frac{{\mathrm{x}}_1+4}{2},\frac{\pm \sqrt{4-{\mathrm{x}}_1^2}}{2})$$$$\mathrm{If}\mathrm{denote}\mathrm{by}(\mathrm{x},\mathrm{y})-\mathrm{the}\mathrm{coordinates}\mathrm{of}$$$$\mathrm{the}\mathrm{point}\mathrm{I}\mathrm{then}\mathrm{we}\mathrm{have}$$$$\{\begin{array}{l}\mathrm{x}=\frac{{\mathrm{x}}_1+4}{2}\left(1\right)\\ \mathrm{y}=\frac{\pm \sqrt{4-{\mathrm{x}}_1^2}}{2}\left(2\right)\end{array}$$$$\mathrm{From}\left(2\right)\mathrm{we}\mathrm{get}{\left(2\mathrm{y}\right)}^2=4-{\mathrm{x}}_1^2$$$$\Rightarrow 4-{\mathrm{x}}_1^2={4\mathrm{y}}^2\Rightarrow {\mathrm{x}}_1^2=4(1-{\mathrm{y}}^2)\left(3\right)$$$$\mathrm{From}\left(1\right)\mathrm{we}\mathrm{get}{\mathrm{x}}_1^2={(2\mathrm{x}-4)}^2$$$$={\left[2(\mathrm{x}-2)\right]}^2=4{(\mathrm{x}-2)}^2\left(4\right)$$$$\mathrm{From}\left(3\right)\mathrm{and}\left(4\right)\mathrm{we}\mathrm{get}$$$$1-{\mathrm{y}}^2={(\mathrm{x}-2)}^2\iff {(\mathrm{x}-2)}^2+{\mathrm{y}}^2=1$$$$\mathrm{This}\mathrm{shows}\mathrm{that}\mathrm{the}\mathrm{point}\mathrm{I}\mathrm{runs}\mathrm{on}$$$$\mathrm{the}\mathrm{circle}\mathrm{with}\mathrm{the}\mathrm{centre}\mathrm{having}$$$$\mathrm{the}\mathrm{cordinates}\mathrm{be}(2,0)\mathrm{and}\mathrm{the}\mathrm{radius}$$$$\mathrm{equal}\mathrm{to}1\mathrm{when}\mathrm{P}\mathrm{remove}\mathrm{on}(\mathrm{O},2)$$

Commented by peter frank last updated on 05/Oct/20

$$\mathrm{thank}\mathrm{you}$$

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