Converting polar equation into rectangular coordinates?

Question

Why does the polar equation r = (2)sin(theta) become x^2 + (y -1)^2 = 1

sahsjing, you are most helpful. However, I don't understand where the +1 comes from. I get: x^2 + y^2 -2y = 0.

What am I missing?

Excellent! Thank you all so much for your help!

Answer 1

Multiply both sides by r,
r^2 = 2rsin(θ)
x^2 + y^2 = 2y, since y = rsin(θ)
x^2 + y^2 - 2y + 1 = 1
=> x^2 + (y -1)^2 = 1
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"However, I don't understand where the +1 comes from. I get: x^2 + y^2 -2y = 0." You add 1 to both sides.

Answer 2

Remember: y = r*sen(theta) and x = r*cos(theta) in the change of coordinates.
I'll write t for theta (shorter).

So;
r = 2*sin(t)
*** multiply both sides by r
r^2=2*r*sin(t)
*** remember 1 = cos^2(t)+sin^2(t)
r^2*(cos^2(t)+sin^2(t))=2*r*sin(t)
*** change coordinates
x^2+y^2=2y
*** add 1 to both sides
x^2+y^2-2y+1=1
X^2+(y-1)^2=1
*** done

Cheers

Answer 3

sin t= y/r so r=2y/r and r^2 =2y.But r^2 = x^2 +y^2
so x^2+y^2-2y=0 which is the same as x^2+(y-1)^2=1

Answer 4

r = 2sinθ,
remember r = √(x2 + y2) and y = rsinθ
multiply both sides of equation above by r
r2 = 2rsinθ, r2 = x2 + y2 and rsinθ= y
substitute x and y values for r and θ terms
x2 + y2 = 2y
rearrange to get all terms on one side
x2 + y2 - 2y = 0
now to get the (y-1)2 term (perfect square, remember)
you must add a 1 to both sides
x2 + (y2 - 2y + 1) = 1
x2 + (y-1)2 = 1

Answer 5

r = sqrt(x^2+y^2)
sintheta = y/r
So sqrt(x^2 + y^2)= 2y/sqrt(x^2+y^2)
x^2+y^2 = 2y
x^2+y^2 -2y = 0
x^2+y^2 -2y +1 =1
x^2+(y-1)^2 = 1