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

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The equation x^2 + y^2 = 4 represents a circle centered at the origin with a radius of 2. To convert it to polar form, substitute x and y with their polar equivalents: x = r cos(θ) and y = r sin(θ). This leads to the equation r^2 cos^2(θ) + r^2 sin^2(θ) = 4, which simplifies using the Pythagorean identity to r^2 = 4. Therefore, the polar form of the equation is r = 2.
Elissa89
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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
 
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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?
 
MarkFL said:
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!
 
Elissa89 said:
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:

$$r=a$$
 
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