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Question Number 13591 by FilupS last updated on 21/May/17
for:  x^2 +(y−1)^2 =1  show that it can be written as:  r=2sin(θ)
$$\mathrm{for}: \\ $$$${x}^{\mathrm{2}} +\left({y}−\mathrm{1}\right)^{\mathrm{2}} =\mathrm{1} \\ $$$$\mathrm{show}\:\mathrm{that}\:\mathrm{it}\:\mathrm{can}\:\mathrm{be}\:\mathrm{written}\:\mathrm{as}: \\ $$$${r}=\mathrm{2sin}\left(\theta\right) \\ $$
Commented by FilupS last updated on 21/May/17
thanks in advanced.  I forgot how to parameterize
$$\mathrm{thanks}\:\mathrm{in}\:\mathrm{advanced}. \\ $$$$\mathrm{I}\:\mathrm{forgot}\:\mathrm{how}\:\mathrm{to}\:\mathrm{parameterize} \\ $$
Answered by ajfour last updated on 21/May/17
r^2 =x^2 +y^2   further  x=rcos θ ;  y=rsin θ  since     x^2 +(y−1)^2 =1  r^2 cos^2 θ+(rsin θ−1)^2 =1  or  r^2 +1−2rsin θ=1  r(r−2sin θ)=0  ⇒  r=0  or/and   r=2sin θ .
$$\mathrm{r}^{\mathrm{2}} =\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} \\ $$$$\mathrm{further}\:\:\mathrm{x}=\mathrm{rcos}\:\theta\:;\:\:\mathrm{y}=\mathrm{rsin}\:\theta \\ $$$$\mathrm{since}\:\:\:\:\:\mathrm{x}^{\mathrm{2}} +\left(\mathrm{y}−\mathrm{1}\right)^{\mathrm{2}} =\mathrm{1} \\ $$$$\mathrm{r}^{\mathrm{2}} \mathrm{cos}\:^{\mathrm{2}} \theta+\left(\mathrm{rsin}\:\theta−\mathrm{1}\right)^{\mathrm{2}} =\mathrm{1} \\ $$$$\mathrm{or} \\ $$$$\mathrm{r}^{\mathrm{2}} +\mathrm{1}−\mathrm{2rsin}\:\theta=\mathrm{1} \\ $$$$\mathrm{r}\left(\mathrm{r}−\mathrm{2sin}\:\theta\right)=\mathrm{0} \\ $$$$\Rightarrow\:\:\mathrm{r}=\mathrm{0}\:\:\mathrm{or}/\mathrm{and}\:\:\:\mathrm{r}=\mathrm{2sin}\:\theta\:. \\ $$

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