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

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

Answered by alephnull last updated on 09/Jan/25

$$=x(2-\cos \left(t\right))=7\cos \left(t\right)-2,y(2-\cos \left(t\right))=4\sqrt{3}\sin \left(t\right)$$$$\mathrm{Rearrange}$$$$2x-x\cos \left(t\right)=7\cos \left(t\right)-2$$$$2y-y\cos \left(t\right)=4\sqrt{3}\sin \left(t\right)$$$$\downarrow$$$$x\cos \left(t\right)=(7\cos \left(t\right)-2)-2x$$$$y\cos \left(t\right)=4\sqrt{3}\sin \left(t\right)-2y$$$$$$$$\mathrm{recall}\mathrm{pythagoras}\mathrm{indentity}$$$$\cos {}^{2}t+\sin {}^{2}t=1$$$$$$$$cos\left(t\right)=\frac{2x+2}{x-7},sin\left(t\right)=\frac{(2-x)y}{4\sqrt{3}}$$$$\mathrm{S}ubstitute$$$${\left(\frac{2x+2}{x-7}\right)}^2+{\left(\frac{(2-x)y}{4\sqrt{3}}\right)}^2=1$$$$\mathrm{simplify}$$$$\frac{{(2x+2)}^2}{{(x-7)}^2}+\frac{{(2-x)}^2{y}^2}{48}=1$$$$=48{(2x+2)}^2+{(2-x)}^2{y}^2{(x-7)}^2=48{(x-7)}^2$$$$$$$$\mathrm{answer}\mathrm{resembles}\mathrm{circle}\mathrm{thing}$$$${(x-h)}^2+{(y-k)}^2=r$$$$$$$$\mathrm{so}\mathrm{it}\mathrm{is}$$$$$$

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

$$x=\frac{7\cos t-2}{2-\cos t}=\frac{12}{2-\cos t}-7$$$$\Rightarrow 2-\cos t=\frac{12}{x+7}$$$$\Rightarrow \cos t=2-\frac{12}{x+7}=\frac{2(x+1)}{x+7}$$$$\Rightarrow \sin t=\frac{y(2-\cos t)}{4\sqrt{3}}=\frac{\sqrt{3}y}{x+7}$$$${\sin }^{2}t+{\cos }^{2}t=1$$$$\Rightarrow {\left(\frac{\sqrt{3}y}{x+7}\right)}^2+4{\left(\frac{x+1}{x+7}\right)}^2=1$$$$\Rightarrow 3y^2+4{(x+1)}^2={(x+7)}^2$$$$\Rightarrow y^2+{(x-1)}^2=4^2$$$$thisisacirclewithradius4and$$$$centerat(1,0)$$

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