Polar equation to rectangular coordinates?

Question

This is my math problem. My math book doesn't provide examples simliar to this problem, which makes it kind of pointless. Nevertheless, the equation is:
****Since I can't type in the Theta symbol, i just wrote theta.

Problem:
r = 2 sin THETA

the answer is x^2 + (y-1)^2 = 1

I just don't understand how to get from the problem to the answer.

Here are some equations that might help:

x = r cos THETA

y = r sin THETA

r^2 = x^2 + y^2

tan THETA = y/x if x doesn't equal to 0

Answer 1

Multiply both sides by r thus:
r^2=2*r[sin(theta)]
so X^2+y^2=2y
Now rearrange.

Answer 2

UNIT CIRCLE
DIAMETER =1
r^2 = x^2 + y^2
ALWAYS 1 = R = Rsq in unit cir, but this appears not unit cir.
------------
r = 2 sin THETA
r is a scalar (?!) , the MAGNITUDE (length) of 2 sin THETA ??
r/2=sin Th
r = sqrt((x^2) + (y ^2))
r^2 = ((x^2) + (y ^2))
x= r sin Th
y = r Cos Th
Sin squared + cos sq = 1


stilll thinkin ....
Is that the WHOLE problem ??
What to find ??
the answer is x^2 + (y-1)^2 = 1
Oh I see, convert a function in Radian to Catrisian

x= r sin Th
y = r Cos Th
You can do this !!

[e-mail]

Answer 3

First, since y = r sin THETA

sinTHETA = y/r, so

2sinTHETA = 2y/r

r = 2y/r

r^2 = 2y (multiply bothsides by r)

...since r^2 = x^2 + y^2, then

x^2 + y^2 = 2y (substitute)

x^2 + y^2 - 2y = 0 (get all terms on the same side)

x^2 + y^2 - 2y + 1 = 0 + 1 [complete the square 1 = (-2/2)^2]

x^2 + (y-1)^2 = 1 [factor]

Answer 4

just expand the answer...
x^2 + y^2 -2y + 1 = 1

solve, substitute, and you'll get sin THETA = r/2,
which is assumed in the question.