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"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.

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3 Answers

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.

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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.

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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 $\frac12(b,a)$.

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