(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 $\alpha$.

My Answer for (i):

$x = r \cdot \cos \alpha$

$y = r \cdot \sin \alpha$

$x^2 + y^2 = R^2$ (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.

I am sorry if this seems like a trivial question, please don't close this question. I haven't done any serious math for the last 2 years, especially geometry. The textbook that he had written (and made us buy) didn't really ease us into the topic but expected us to already know all these formulas, I only managed to do these much after 6 hours of googling and reading up on functions, polar equations, and so on. Please please please help, thanks!!

  • $\begingroup$ 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) $\endgroup$ Aug 13, 2020 at 5:37
  • $\begingroup$ Thanks Ninad, does that mean that the domain range is (0, 2$\pi$), which means all the angles from 0 to 360 degrees? $\endgroup$ Aug 13, 2020 at 6:03
  • $\begingroup$ Yes it does, and zkutch has an answer for the other one below. $\endgroup$ Aug 13, 2020 at 6:04

1 Answer 1


For first think about period for trigonometry functions.

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.