Question Number 209131 by mahdipoor last updated on 02/Jul/24
$${prove}\:: \\ $$$${curve}\:\begin{cases}{{x}\left({t}\right)=\frac{{a}+{r}.{cos}\left({t}\right)}{{a}^{\mathrm{2}} +{r}^{\mathrm{2}} +\mathrm{2}{ar}.{cos}\left({t}\right)}}\\{{y}\left({t}\right)=\frac{{r}.{sin}\left({t}\right)}{{a}^{\mathrm{2}} +{r}^{\mathrm{2}} +\mathrm{2}{ar}.{cos}\left({t}\right)}}\end{cases}\:\:\:\:\mathrm{0}\leqslant{t}\leqslant\mathrm{2}\pi \\ $$$${is}\:{circle}\:,\:{find}\:{center}\:\&\:{radius} \\ $$
Answered by Frix last updated on 02/Jul/24
$$\begin{cases}{\mathrm{cos}\:{t}\:=\frac{{a}−\left({a}^{\mathrm{2}} +{r}^{\mathrm{2}} \right){x}}{\left(\mathrm{2}{ax}−\mathrm{1}\right){r}}}\\{{y}=\pm\frac{{r}\sqrt{\mathrm{1}−\mathrm{cos}^{\mathrm{2}} \:{t}}}{{a}^{\mathrm{2}} +{r}^{\mathrm{2}} +\mathrm{2}{ar}\mathrm{cos}\:{t}}}\end{cases} \\ $$$$\Rightarrow \\ $$$$\left({y}−\frac{{r}\sqrt{\mathrm{1}−\mathrm{cos}^{\mathrm{2}} \:{t}}}{{a}^{\mathrm{2}} +{r}^{\mathrm{2}} +\mathrm{2}{ar}\mathrm{cos}\:{t}}\right)\left({y}+\frac{{r}\sqrt{\mathrm{1}−\mathrm{cos}^{\mathrm{2}} \:{t}}}{{a}^{\mathrm{2}} +{r}^{\mathrm{2}} +\mathrm{2}{ar}\mathrm{cos}\:{t}}\right)=\mathrm{0} \\ $$$$\Leftrightarrow \\ $$$$\left({x}−\frac{{a}}{{a}^{\mathrm{2}} −{r}^{\mathrm{2}} }\right)^{\mathrm{2}} +{y}^{\mathrm{2}} =\left(\frac{{r}}{{a}^{\mathrm{2}} −{r}^{\mathrm{2}} }\right)^{\mathrm{2}} \\ $$$$\mathrm{Center}=\begin{pmatrix}{\frac{{a}}{{a}^{\mathrm{2}} −{r}^{\mathrm{2}} }}\\{\mathrm{0}}\end{pmatrix}\:,\:\mathrm{Radius}=\frac{{r}}{{a}^{\mathrm{2}} −{r}^{\mathrm{2}} } \\ $$