prove : curve { ((x(t)=((a+r.cos(t))/(a^2 +r^2 +2ar.cos(t))))),((y(t)=((r.sin(t))/(a^2 +r^2 +2ar.cos(t))))) :} 0≤t≤2π is circle , find center & radius


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Question Number 209131 by mahdipoor last updated on 02/Jul/24
$prove : curve { ((x(t)=((a+r.cos(t))/(a^2 +r^2 +2ar.cos(t))))),((y(t)=((r.sin(t))/(a^2 +r^2 +2ar.cos(t))))) :} 0≤t≤2π is circle , find center & radius$
$p r o v e :$ $c u r v e {\begin{cases} x (t) = \frac{a + r . c o s (t)}{a^{2} + r^{2} + 2 a r . c o s (t)} \\ y (t) = \frac{r . s i n (t)}{a^{2} + r^{2} + 2 a r . c o s (t)} \end{cases} 0 ⩽ t ⩽ 2 π$ $i s c i r c l e, f i n d c e n t e r & r a d i u s$
	Answered by Frix last updated on 02/Jul/24
	${ ((cos t =((a−(a^2 +r^2 )x)/((2ax−1)r)))),((y=±((r(√(1−cos^2 t)))/(a^2 +r^2 +2arcos t)))) :} ⇒ (y−((r(√(1−cos^2 t)))/(a^2 +r^2 +2arcos t)))(y+((r(√(1−cos^2 t)))/(a^2 +r^2 +2arcos t)))=0 ⇔ (x−(a/(a^2 −r^2 )))^2 +y^2 =((r/(a^2 −r^2 )))^2 Center= (((a/(a^2 −r^2 ))),(0) ) , Radius=(r/(a^2 −r^2 ))$
	${\begin{cases} \cos t = \frac{a - (a^{2} + r^{2}) x}{(2 a x - 1) r} \\ y = \pm \frac{r \sqrt{1 - \cos^{2} t}}{a^{2} + r^{2} + 2 a r \cos t} \end{cases}$ $\Rightarrow$ $(y - \frac{r \sqrt{1 - \cos^{2} t}}{a^{2} + r^{2} + 2 a r \cos t}) (y + \frac{r \sqrt{1 - \cos^{2} t}}{a^{2} + r^{2} + 2 a r \cos t}) = 0$ $\Leftrightarrow$ ${(x - \frac{a}{a^{2} - r^{2}})}^{2} + y^{2} = {(\frac{r}{a^{2} - r^{2}})}^{2}$ $Center = (\begin{matrix} \frac{a}{a^{2} - r^{2}} \\ 0 \end{matrix}), Radius = \frac{r}{a^{2} - r^{2}}$
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