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-2pi-is-circle-find-center-amp-radius-

Question Number 209131 by mahdipoor last updated on 02/Jul/24

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

{\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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