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Question Number 13591 by FilupS last updated on 21/May/17

$$\mathrm{for}:$$$$x^2+{(y-1)}^2=1$$$$\mathrm{show}\mathrm{that}\mathrm{it}\mathrm{can}\mathrm{be}\mathrm{written}\mathrm{as}:$$$$r=2\sin \left(\theta \right)$$

Commented by FilupS last updated on 21/May/17

$$\mathrm{thanks}\mathrm{in}\mathrm{advanced}.$$$$\mathrm{I}\mathrm{forgot}\mathrm{how}\mathrm{to}\mathrm{parameterize}$$

Answered by ajfour last updated on 21/May/17

$${\mathrm{r}}^2={\mathrm{x}}^2+{\mathrm{y}}^2$$$$\mathrm{further}\mathrm{x}=\mathrm{rcos}\theta ;\mathrm{y}=\mathrm{rsin}\theta$$$$\mathrm{since}{\mathrm{x}}^2+{(\mathrm{y}-1)}^2=1$$$${\mathrm{r}}^2\cos {}^2\theta +{(\mathrm{rsin}\theta -1)}^2=1$$$$\mathrm{or}$$$${\mathrm{r}}^2+1-2\mathrm{rsin}\theta =1$$$$\mathrm{r}(\mathrm{r}-2\sin \theta )=0$$$$\Rightarrow \mathrm{r}=0\mathrm{or}/\mathrm{and}\mathrm{r}=2\sin \theta .$$

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