Converting x^2+y^2=4 to polar form

[Elissa89]

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]

You need to also square the trig. functions:

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

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

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

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

 

[Elissa89]

You need to also square the trig. functions:

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

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

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

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

got it! thanks!

 

[MarkFL]

got it! thanks!

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:

$$x^2+y^2=a^2$$ where \(0<a\)

Has the polar equation:

$$r=a$$