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I'm reviewing my notes and I came upon this. It says that $$ \begin{align} r \sin \theta &= 5 \tag{horizontal line} \\ r &= \dfrac{3 \pi}{5} \tag{diagonal line} \end{align}$$ How is this possible, when $\dfrac{3 \pi}{5}$ results to a constant (therefore a straight line)?

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You probably mean theta = 3 pi / 5. – FiveLemon May 9 '13 at 1:27
$r=\frac{3\pi}{5}$ should be a circle of radius $\frac{3\pi}{5}$ as opposed to a diagonal line. – Maazul May 9 '13 at 1:31
up vote 7 down vote accepted

The equation $r=k$ is the equation of a circle, centre the origin, radius $k$. So it would be a circle of radius $\frac{3\pi}{5}$. Odd radius!

You presumably meant $\theta=\frac{3\pi}{5}$, which (depending on whether we don't allow negative $r$ or do) is a half-line or a full line.

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Ah. Now I remember our instructor retracting her answer to this particular question. She did say it was a circle. Thanks! – Gannicus May 9 '13 at 1:36

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