Prove-that-x-7-cos-t-2-2-cos-t-y-4-3-sin-t-2-cos-t-is-a-circle-

Question Number 215520 by CrispyXYZ last updated on 09/Jan/25

Prove that {\begin{cases} x = \frac{7 \cos t - 2}{2 - \cos t} \\ y = \frac{4 \sqrt{3} \sin t}{2 - \cos t} \end{cases} is a circle .

Answered by alephnull last updated on 09/Jan/25

= x (2 - \cos (t)) = 7 \cos (t) - 2, y (2 - \cos (t)) = 4 \sqrt{3} \sin (t)

Rearrange

2 x - x \cos (t) = 7 \cos (t) - 2

2 y - y \cos (t) = 4 \sqrt{3} \sin (t)

↓

x \cos (t) = (7 \cos (t) - 2) - 2 x

y \cos (t) = 4 \sqrt{3} \sin (t) - 2 y

recall pythagoras indentity

\cos^{2} t + \sin^{2} t = 1

c o s (t) = \frac{2 x + 2}{x - 7}, s i n (t) = \frac{(2 - x) y}{4 \sqrt{3}}

S u b s t i t u t e

{(\frac{2 x + 2}{x - 7})}^{2} + {(\frac{(2 - x) y}{4 \sqrt{3}})}^{2} = 1

simplify

\frac{{(2 x + 2)}^{2}}{{(x - 7)}^{2}} + \frac{{(2 - x)}^{2} y^{2}}{48} = 1

= 48 {(2 x + 2)}^{2} + {(2 - x)}^{2} y^{2} {(x - 7)}^{2} = 48 {(x - 7)}^{2}

answer resembles circle thing

{(x - h)}^{2} + {(y - k)}^{2} = r

so it is

Answered by mr W last updated on 10/Jan/25

x = \frac{7 \cos t - 2}{2 - \cos t} = \frac{12}{2 - \cos t} - 7

\Rightarrow 2 - \cos t = \frac{12}{x + 7}

\Rightarrow \cos t = 2 - \frac{12}{x + 7} = \frac{2 (x + 1)}{x + 7}

\Rightarrow \sin t = \frac{y (2 - \cos t)}{4 \sqrt{3}} = \frac{\sqrt{3} y}{x + 7}

\sin^{2} t + \cos^{2} t = 1

\Rightarrow {(\frac{\sqrt{3} y}{x + 7})}^{2} + 4 {(\frac{x + 1}{x + 7})}^{2} = 1

\Rightarrow 3 y^{2} + 4 {(x + 1)}^{2} = {(x + 7)}^{2}

\Rightarrow y^{2} + {(x - 1)}^{2} = 4^{2}

t h i s i s a c i r c l e w i t h r a d i u s 4 a n d

c e n t e r a t (1, 0)

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