Graphical practice G3: polar pairs, a circle and the origin

Separate a coordinate pair from the point it represents

With nonnegative radius, polar coordinates use the map

$$\Phi:[0,\infty)\times\mathbb R\longrightarrow\mathbb R^2,\qquad \Phi(r,\theta)=(r\cos\theta,r\sin\theta)$$

Angles are in radians. At $r>0$, adding an integer multiple of $2\pi$ to the angle leaves the represented point unchanged. At $r=0$, every angle represents the origin. This describes the coordinate map; a particular polar equation can impose further restrictions on which pairs are solutions.

If $S$ is the set of allowed coordinate pairs satisfying an equation, its planar locus is $\Phi(S)=\{\Phi(r,\theta):(r,\theta)\in S\}$. Membership in this image means that at least one allowed pair represents the point. It does not require every representation of that point to satisfy the equation. A later restriction to $-\pi\le\theta\le\pi$ changes the allowed pair domain and must be applied before listing its angle solutions.

Prove the point set in both directions

Replacing $r^2$ by $x^2+y^2$ and $r\cos\theta$ by $x$ proves a rectangular consequence of the polar equation. To prove equality of loci, one must also start with a rectangular point and find an allowed polar pair satisfying the original equation. For a nonzero point choose $r=\sqrt{x^2+y^2}>0$; division by $r$ is then legitimate. At the origin division is unavailable, so check a permitted angle separately.

This is a question about logical reversibility, not the appearance of the graph. Equal axis scales let the diagram support the geometric interpretation of centre and radius; the two inclusions establish the exact point set. The following four questions keep these two tasks distinct.

Read the circle and answer the four questions

Checkpoint

Graphical practice G3: polar pairs, a circle and the origin

Let $r,x,y$ be numerical lengths measured in centimetres, and let $\theta$ be in radians. Use nonnegative polar radii and

$$x=r\cos\theta,\qquad y=r\sin\theta.$$

Consider the polar equation

$$r=4\cos\theta,\qquad r\ge0.$$

The figure uses equal scales on the two rectangular axes. The ray through $P$ has direction $\theta=\pi/3$.

1. Convert the polar equation to a rectangular equation. Verify the centre and radius read from the graph. Prove the converse and check the origin separately, so that the full point set is established.
2. When $-\pi\le\theta\le\pi$, which angles permit $r>0$? Which angles represent the origin while satisfying the polar equation?
3. Find $r$ and the rectangular coordinates of $P$ exactly. A whole ray intersects the rectangular circle at the origin as well as at $P$. Does the pair $(r,\theta)=(0,\pi/3)$ satisfy the original polar equation? Explain the distinction.
4. A student cancels $r$ in $r^2=4r\cos\theta$. Does this preserve all polar-coordinate pairs? Does it preserve the rectangular point set in this example?

Complete solutions: pairs, points and the zero branch

Solution · Graphical practice G3: polar pairs, a circle and the origin

1. Multiplying by $r$ gives $r^2=4r\cos\theta$, hence

   $$x^2+y^2=4x\iff(x-2)^2+y^2=4.$$

   The centre is $(2,0)$ centimetres and the radius is two centimetres, matching the undistorted circle in the figure. Multiplication alone proves only one implication. Conversely, take a nonzero point satisfying the rectangular equation, choose its nonnegative radius $r=\sqrt{x^2+y^2}>0$ and a corresponding angle. Then $r^2=4r\cos\theta$, and division by this positive radius recovers $r=4\cos\theta$. The origin also belongs to the polar locus: $r=0$, $\theta=\pi/2$ satisfies the equation. Thus no rectangular point was lost or added.

2. For a strictly positive radius, $\cos\theta>0$, so within the stated angle window we require $-\pi/2<\theta<\pi/2$. The origin has $r=0$ and the equation then requires $\cos\theta=0$, giving $\theta=-\pi/2$ or $\pi/2$ in this window. Each pair describes the same origin. Negative cosine values would require negative radii and are not allowed under this convention. Letting $\theta$ vary from $-\pi/2$ to $\pi/2$ traces the full circle, with the origin represented at both ends.

3. At $\theta=\pi/3$, $r=4(1/2)=2$. Thus

   $$P=(2\cos(\pi/3),2\sin(\pi/3))=(1,\sqrt3)\quad\text{centimetres}.$$

   Its squared displacement from the centre is $(1-2)^2+(\sqrt3)^2=4$, as required. But $(0,\pi/3)$ does not satisfy $r=4\cos\theta$, since it would assert $0=2$. A rectangular point can have several polar representations; membership of a point in a locus means that at least one permitted representation satisfies its equation, not that every representation does. The origin lies on the ray and circle as a rectangular point, while the polar equation represents it at different angles.

4. The multiplied equation holds at $r=0$ for every angle, whereas the cancelled equation holds there only when $\cos\theta=0$. Cancelling therefore removes many polar-coordinate pairs and is not reversible on the full pair domain. Nevertheless, in this example every removed pair represents the same origin, and the origin is still represented by a surviving pair. At positive radius the division is reversible. The two equations thus have the same rectangular point set even though their polar-pair solution sets differ.

When cancellation changes pairs but preserves points

In this example, multiplying $r=4\cos\theta$ by $r$ adds all pairs with $r=0$ that were not already solutions. Every added pair maps to the origin, and that point already has an original solution pair, such as $(0,\pi/2)$. Hence the image point set is unchanged even though the pair set grows. This conclusion depends on checking the zero-radius branch.

It is not a general license to multiply or divide polar equations without checking. For example, $r=1$ with $r\ge0$ describes the unit circle; multiplying by $r$ gives $r^2=r$ and adds the origin to the planar locus. In either direction, record which pairs are lost or added and then examine their images under $\Phi$.