find polar equation

[Guest]

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

 

[stapel]

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

Eliz.

 

[pka]

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>.

 

[soroban]

Hello, Beckie!

Convert: x² + y² = 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: x² + y² = r² ?

but I don't see the relation of the result? . . . you don't?

. . . . . x² + y² . = . 2 c x
. . . . . . . .↓ . . . . . . . . . ↓
. . . . . . . .r² . . .= .2 c (r cosθ)


I don't see how it can be any easier . . .

 

[Guest]

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