# Define polar coordinates of circle at origin and circle with radius $R$.

Question:

(i) Define in polar coordinates $$r = f(\alpha)$$ the origin-centred circle with radius $$R$$. Specify the domain range for the polar coordinate $$\alpha$$.

(ii) Define in polar coordinates $$r = f(\alpha)$$ a circle with radius $$R$$ and the centre at the Cartesian coordinates $$(R, 0)$$. Specify the domain range for the polar coordinate $$\alpha$$.

$$x = r \cdot \cos \alpha$$

$$y = r \cdot \sin \alpha$$

$$x^2 + y^2 = R^2$$ (equation for circle, radius R centered at origin)

$$r^2(\cos^2 \alpha + \sin^2 \alpha) = R^2$$

$$r^2 = R^2$$

$$r = R$$

The domain range of $$\alpha$$ is $$(-\infty, \infty)$$?

Is my answer correct? I don't think it's correct because my RHS doesn't have $$\alpha$$ at all. I've learnt that domain is the set valid input values and range is the set valid output values. However, I don't understand what the question means by "domain range" of $$\alpha$$?

For part (ii) I don't understand how do you shift a circle centered at origin to become the radius.

• For the first one you're right that $\alpha$ has no restriction, but the question is really asking what is the smallest set of angles you could get away with, but still retrieve the whole circle (so $(-\infty,\infty)$ is definitely overkill because you repeat the circle many, many times) Aug 13, 2020 at 5:37
• Thanks Ninad, does that mean that the domain range is (0, 2$\pi$), which means all the angles from 0 to 360 degrees? Aug 13, 2020 at 6:03
• Yes it does, and zkutch has an answer for the other one below. Aug 13, 2020 at 6:04

For second you have $$(x-R)^2+y^2=R^2$$ circle. Inserting polar coordinates gives $$(x-R)^2+y^2=R^2 \Leftrightarrow r-2\cos \alpha=0$$ From here you obtain also restrictions for $$\alpha$$: $$\cos \alpha \geqslant 0$$ and its enough to find one continuous segment for it.
And, at last little gift for second: consider $$x-R = r \cdot \cos \alpha$$ and $$y = r \cdot \sin \alpha$$. This is little "extended" polar coordinates, but sometimes such shifting center creates big simplicity.
If I can advise something, then, it is to try to "look at world with polar eyes": disk bounded by circle in polar coordinates becomes rectangle. Disk with shifted center, as in your 2nd example, becomes area under $$\cos$$ function, but if you use "gift", then it again will be rectangle. Try to draw every figure you work with in polar coordinates so, if they are simple cartesian $$(r, \alpha)$$. Same world with another eyes.