Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

"Show that the polar equation $r = a \sin(t) + b \cos(t)$, where $ab \neq 0$, represents a circle, and find its center and radius."

I don't need the answer - I just need a hint to nudge me on. I've tried to tackle it by investigating tangents, completing squares, force-fitting into the formula of a circle, etc., but I can't seem to arrive at the indisputable conclusion that it must be a circle.

This is not homework, by the way. It comes from a Calculus textbook that I am studying on my own. I understand you have no reason to believe me, but if I were taking a real course, I'd bring this to my TA.

share|cite|improve this question
up vote 3 down vote accepted

Multiply either side by $r$, use $r^2=x^2+y^2, r\sin t=y,r\cos t=x$

But we don't need $ab\ne 0$

If $a=0,r=b\cos t\implies r^2=b r\cos t\implies x^2+y^2-bx=0$ $\implies (x-\frac b 2)^2+y^2=(\frac b 2)^2$ is a circle.

If both $a,b$ are $0, r=0\implies x^2+y^2=0$ which is point-circle.

share|cite|improve this answer

Hint: Use the relations $\sin{t} = \frac{y}{r}$, $\cos{t} = \frac{x}{r}$ and $r = \sqrt{x^2 + y^2}$ to find a cartesian equation for the circle.

share|cite|improve this answer

Hint 1:

Compute $x$ and $y$: $$ \begin{array}{lll} x&=r\cos(t)&=a\sin(t)\cos(t)+b\cos^2(t)&=\frac{a}{2}\sin(2t)+\frac{b}{2}(1+\cos(2t))\\ y&=r\sin(t)&=a\sin^2(t)+b\sin(t)\cos(t)&=\frac{a}{2}(1-\cos(2t))+\frac{b}{2}\sin(2t) \end{array} $$

Hint 2:

Therefore, if we set $(b,a)=\sqrt{a^2+b^2}(\cos(\theta),\sin(\theta))$, we get $$ (x,y)=\tfrac12(b,a)+\tfrac12\sqrt{a^2+b^2}\;(\cos(2t-\theta),\sin(2t-\theta)) $$

Answer: (mouse over to view)

Thus, the curve is a circle of radius $\frac12\sqrt{a^2+b^2}$ centered at $\frac12(b,a)$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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