Question

Write r^2-2rsintheta=0 in rectangular form.

Answer 1

I'll give you a hint... y = rsinθ and x = rcosθ

Answer 2

yeah its r\[r^{2}-2rsin \theta=0\]

Answer 3

I got up to x^2+y^2-2y=0, but I think its not complete

Answer 4

ok, can you show me what I should do next

Answer 5

...I think I'm stuck lol

Answer 6

XD

Answer 7

WAIT NO I GOT IT

Answer 8

Give me a few minutes xD

Answer 9

ok lol

Answer 10

okay I think I have it down now\[r^{2}-2r\sin{\theta}=0\]\[r^{2}=2r\sin{\theta}\]As rsinθ = y, \(r^{2}=2y\). so the equation \(r^{2}=x^{2}+y^{2}\) becomes \(2y=x^{2}+y^{2}\)

Answer 11

Are you following me? xD

Answer 12

yeah, I see it, do you think that's it?

Answer 13

Oh.
No, the equation \(x^{2}+y^{2}=r^{2}\) only works if your circle's center is the origin. Here the circle's center is (h,k) or \((x-h)^{2}+(y-k)^{2}=r^{2}\)
I don't know if it's clear just yet but there's a k value because of the 2y

Answer 14

ok so what should i do, lol

Answer 15

So if we move 2y over to the left side we get \(0=x^{2}+y^{2}-2y\). Then you can complete the square for the y-variable

Answer 16

Do you know how to complete the square?

Answer 17

I don't think I do, sorry

Answer 18

It's okay, that's fine.
Completing the square goes something like this - if we have equation \(x^{2}+bx=0\) we can add \(c=\left(\frac{b}{2}\right)^{2}\) to both sides to get \(x^{2}+bx+c=c\) (Actually, you're supposed to make the other side 0 but that's another type of problem)

Answer 19

So here we have \(x^{2}+y^{2}-2y=0\), and you can perform the completing the square action on the y-values

Answer 20

\[x^{2}+(y^{2}-2y+\left(\frac{2}{2}\right)^{2})=0+\left(\frac{2}{2}\right)^{2}\]

Answer 21

I think you can do it from here... Just simplify everything ^_^

Answer 22

ok, Thanks!

Answer 23

But I'll be here if you need me to explain anything else :D

Answer 24

No problem (:
\(\text{Happy OpenStudying!!}\)

Answer 25

You Too, XD

Answer 26

;)