# Thread: Converting from Polar form to Rectangular form

1. ## Converting from Polar form to Rectangular form

If you have the time,

So i want to convert this polar equation to a rectangular form.
Problem1: r= 4cosθ - 4sinθ

I understand the simple conversions like this one:
r= 3secθ
r/secθ = 3 (divided secθ)
rcosθ = 3 (1/secθ = cosθ)
x = 3 : rcosθ = x (coordinate conversion equations)

but i dont know how to begin with the problem 1

Here are other coordinate conversion equations that might help:
y = rsinθ
r^2 = x^2 + y^2
tanθ = y/x

thankyou!

2. Originally Posted by yeunju
If you have the time,

So i want to convert this polar equation to a rectangular form.
Problem1: r= 4cosθ - 4sinθ
Hint: multiply through by r, you get

$r^2 = 4r \cos \theta - 4 r \sin \theta$

now what?

3. Oh, with that helpful hint
i'm thinking to use the x=rcosθ and y=rsinθ and r^2 = x^2+y^2
so: x^2 + y^2 = 4x - 4y
then maybe x^2 -4x + y^2 + 4y = 0

I remember an example with my teacher using complete the square? but it was on an equation like x^2 + y^2 = 4x (subtract 4x)
so then it ended up with x^2 -4x +4 +y^2 = 4

so could i do that with both x and y? maybe..
or am i going the wrong direction?

4. Yes, rewrite it as $x^2 -4x +4 +y^2 +4y+4 = 8$.

Complete the square for both x and y to get:

Spoiler:
$(x-2)^2+(y+2)^2=8$

Voila! It's a circle of radius $2\sqrt{2}$ centered at $(2,-2)$.