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

Elissa89

Member
Oct 19, 2017
52
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

Administrator
Staff member
Feb 24, 2012
13,775
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
Oct 19, 2017
52
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

Administrator
Staff member
Feb 24, 2012
13,775
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:

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

Has the polar equation:

\(\displaystyle r=a\)