Question 121066

Using the conversion formulas that convert polar to rectangular, we have 



*[Tex \LARGE x=r\cos(\theta)]


and 


*[Tex \LARGE r=\sqrt{x^{2}+y^{2}}]





*[Tex \LARGE r^{2}=x^{2}+y^{2}] Now square both sides of the second formula



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Now let's manipulate *[Tex \LARGE r=\cos(\theta)]:



*[Tex \LARGE r^2=r\cos(\theta)] Multiply both sides by r



*[Tex \LARGE x^{2}+y^{2}=r\cos(\theta)] Replace *[Tex \LARGE r^{2}] with *[Tex \LARGE x^{2}+y^{2}]. Remember, *[Tex \LARGE r^{2}=x^{2}+y^{2}] 




*[Tex \LARGE x^{2}+y^{2}=x] Replace *[Tex \LARGE r\cos(\theta)] with *[Tex \LARGE x]. Remember, *[Tex \LARGE x=r\cos(\theta)]




So the polar equation *[Tex \LARGE r=\cos(\theta)] converts to *[Tex \LARGE x^{2}+y^{2}=x]


So the answer is A)



Check:

You can graph the original equation *[Tex \LARGE r=\cos(\theta)] and *[Tex \LARGE x^{2}+y^{2}=x]. You'll see that they trace out the same circle.