# convert another equation x^2+y^2=4 to polar form

#### Elissa89

##### Member
x^2+y^2=4

I have so far:

(r^2)cos^(theta)+(r^2)sin(theta)=4

Idk what I'm supposed to do from here

#### MarkFL

Staff member
You need to also square the trig. functions:

$$\displaystyle x^2+y^2=4$$

$$\displaystyle (r\cos(\theta))^2+(r\sin(\theta))^2=4$$

$$\displaystyle r^2\cos^2(\theta)+r^2\sin^2(\theta)=4$$

Factor the LHS...what do you have...is there a trig. identity you can apply?

#### Elissa89

##### Member
You need to also square the trig. functions:

$$\displaystyle x^2+y^2=4$$

$$\displaystyle (r\cos(\theta))^2+(r\sin(\theta))^2=4$$

$$\displaystyle r^2\cos^2(\theta)+r^2\sin^2(\theta)=4$$

Factor the LHS...what do you have...is there a trig. identity you can apply?
got it! thanks!

#### MarkFL

We see we have a circle centered at the origin, and in polar coordinates, that's simply a constant value for $$r$$. The Cartesian equation:
$$\displaystyle x^2+y^2=a^2$$ where $$0<a$$
$$\displaystyle r=a$$