find polar equation

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Convert: x^2 + y^2 = 2cx to a polar equation

I sort of understand how to do this but the answer in the book is not what I'm getting.

The book says go from: x^2 + y^2 = 2cx ....... to r^2 = 2cr cos@

I do not understand how this went to that

I know there are some formulas I can use such as:

x = rcos@ and y = rsin@ but I don't see the relation of the result?

Thank you so much for helping me

Take care,
Beckie
 
This one works just like the other one you posted, the only difference being the "2c" replacing the "-8".

Eliz.
 
You know that x = rcos(Θ) and y = rsin(Θ).
Also cos<SUP>2</SUP>(Θ)+sin<SUP>2</SUP>(Θ)=1.
So r<SUP>2</SUP>= r<SUP>2</SUP>(1)= r<SUP>2</SUP>( cos<SUP>2</SUP>(Θ)+sin<SUP>2</SUP>(Θ))=[rcos(Θ)]<SUP>2</SUP>+[rsin(Θ)]<SUP>2</SUP>.
 
Hello, Beckie!

Convert: x<sup>2</sup> + y<sup>2</sup> = 2cx to a polar equation

I know there are some formulas I can use such as: . . . x = r cosθ and y = r sinθ
. . How about: x<sup>2</sup> + y<sup>2</sup> = r<sup>2</sup> ?

but I don't see the relation of the result? . . . you don't?
. . . . . x<sup>2</sup> + y<sup>2</sup> . = . 2 c x
. . . . . . . . . . . . . . . . .
. . . . . . . .r<sup>2</sup> . . .= .2 c (r cosθ)


I don't see how it can be any easier . . .
 
The reason it isn't clear is because it's really strange to me to evaluate an expression this way. Usually I would plug x and y in and get my answer. I think it's confusing because maybe x^2 + y^2 is a function of r^2 or something like that. Maybe that is why it is so confusing to me. But I am understanding it more now that I found out how to do all the steps that lead me to the right answer. I have to have everything connect. I can't just jump to the right answer otherwise it confuses me.

I hope that helps why it's confusing to me.

Take care,
Beckie
 
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