Precalculus

Polar circles and symmetry tests

Which polar equations draw a circle, how do you sketch one without plotting a hundred points, and how can you tell from the equation alone that its graph is the same on both sides of a line?

You need

Converting between polar and rectangularPrecalculus Plotting polar pointsPrecalculus Line symmetryGrade 4

Where you are 3 in Polar equations and their graphs

Trigonometry7

Polar coordinates and vectors10

Polar equations and their graphs3

Limacons and cardioids

Polar circles and symmetry tests

Rose curves

When you reach for it

A polar equation r = f(\theta) is graphed by choosing directions \theta and walking out the distance r that the equation gives (Plotting polar points). That works for any equation but takes many points. Four families of equations are worth knowing on sight, because their graphs are a circle or a line and can be drawn at once:

r = a, \qquad \theta = c, \qquad r = a\cos\theta, \qquad r = a\sin\theta

Each can be confirmed by converting to rectangular form (Converting an equation between the two systems). For other polar curves, three quick symmetry tests on the equation tell you which half, or which quarter, of the graph is enough to compute.

Throughout, the pole is the origin, the polar axis is the positive x-axis, a \ne 0, and r may be negative: (r, \theta) with r < 0 is the point |r| units out in the direction \theta + \pi.

The procedure

Constant r, constant theta

The graph of r = a is every point at distance |a| from the pole: the circle of radius |a| centered at the pole, x^2 + y^2 = a^2. A negative a gives the same circle, since (a, \theta) and (-a, \theta + \pi) name the same point.

The graph of \theta = c is every point (r, c) with r any real number: the ray at angle c (r > 0), the opposite ray (r < 0) and the pole. Together they make the whole line through the pole at angle c.

The circles r = a cos theta and r = a sin theta

Multiply r = a\cos\theta by r and use r^2 = x^2 + y^2 and x = r\cos\theta:

\begin{aligned} r^2 &= ar\cos\theta \\ x^2 + y^2 &= ax \\ \left(x - \tfrac{a}{2}\right)^2 + y^2 &= \tfrac{a^2}{4} \end{aligned}

the last line by completing the square. Multiplying by r adds nothing here, because the original already reaches the pole: r = 0 at \displaystyle \theta = \frac{\pi}{2}. So the graph is the circle with center \displaystyle \left(\frac{a}{2}, 0\right) and radius \displaystyle \frac{|a|}{2}. It passes through the pole, and its diameter from the pole runs along the polar axis line to the point (a, 0). In the same way r = a\sin\theta becomes x^2 + y^2 = ay, the circle with center \displaystyle \left(0, \frac{a}{2}\right) and radius \displaystyle \frac{|a|}{2}.

Figure 1The sign of a chooses the side; cosine puts the circle on the polar axis line, sine on the line \displaystyle \theta = \frac{\pi}{2}.

There is also a reason you can see. Take a > 0, let A be the point (a, 0), and let P be a point of the circle with diameter OA, in the direction \theta from the pole. An angle inscribed in a semicircle is a right angle, so triangle OPA has its right angle at P, and in that right triangle OP = OA\cos\theta. That is r = a\cos\theta.

Figure 2The angle at P is a right angle, so r = 6\cos\theta. This is why the graph is a circle.

Why the circle is traced once on [0, pi)

As \theta runs from 0 up to (not including) \pi, the line through the pole at angle \theta takes every direction exactly once. Each such line meets the circle r = a\cos\theta at the pole and at one other point, except the tangent line \displaystyle \theta = \frac{\pi}{2}, which meets it only at the pole. The equation gives one point on each line, so the point moves once around the whole circle. On [\pi, 2\pi) it goes around again, because

a\cos(\theta + \pi) = -a\cos\theta

and (-r, \theta + \pi) is the same point as (r, \theta). So over [0, 2\pi) the circle is traced twice. The same holds for r = a\sin\theta, since \sin(\theta + \pi) = -\sin\theta too.

The three symmetry tests

A graph has Line symmetry about a line when reflecting it across that line gives the same graph. A polar graph is checked against three mirrors: the polar axis (the x-axis), the line \displaystyle \theta = \frac{\pi}{2} (the y-axis), and the pole (a half turn about the origin, the same moves that relate points of the unit circle). Each mirror image of (r, \theta) has two convenient names:

Mirror	Image of (r, \theta)	Another name for it
polar axis	(r, -\theta)	(-r, \pi - \theta)
the line \displaystyle \theta = \frac{\pi}{2}	(r, \pi - \theta)	(-r, -\theta)
the pole	(-r, \theta)	(r, \theta + \pi)

Figure 3One name for each mirror image of (r, \theta). Each image has other names too.

The test. Replace (r, \theta) in the equation by one of the names in the mirror’s row. If the new equation simplifies to the original, the graph has that symmetry. The reason: when (r, \theta) satisfies the equation, so does the mirror name, and so the mirror image of every point of the graph is on the graph.

A failed test proves nothing. A point is on a polar graph when some name of it satisfies the equation. If the replacement does not give back the original equation, the mirror image might still lie on the graph through a different name. So try both names in the row; and if both fail, the symmetry is still possible. To show a graph is not symmetric, find one point of the graph whose mirror image is not on it, or convert and look.

Worked examples

Example 1

A constant r and a constant theta

Graph r = 4 and \displaystyle \theta = \frac{3\pi}{4} on the same polar grid, and give each in rectangular form.

Solution

Every point of r = 4 is 4 units from the pole, in any direction, so the graph is the circle of radius 4 about the pole: x^2 + y^2 = 16.

The points of \displaystyle \theta = \frac{3\pi}{4} are \displaystyle (r, \frac{3\pi}{4}) for every real r. Positive r walks into quadrant II, negative r walks backward into quadrant IV, and r = 0 is the pole, so the graph is the whole line. Its slope is \displaystyle \tan\frac{3\pi}{4} = -1, so it is y = -x.

Figure 4Example 1. A constant r is a circle about the pole; a constant \theta is a line through it.

Example 2

A table for r = 6 cos theta

Graph r = 6\cos\theta from a table of values on [0, \pi], and confirm the shape by converting.

Solution

Exact values of cosine at the standard angles give the table. The last column turns each (r, \theta) into (x, y) with x = r\cos\theta, y = r\sin\theta.

\theta	r = 6\cos\theta	(x, y)
0	6	(6, 0)
\displaystyle \frac{\pi}{6}	\displaystyle 3\sqrt{3} \approx 5.20	\displaystyle \left(\frac{9}{2}, \frac{3\sqrt{3}}{2}\right)
\displaystyle \frac{\pi}{4}	\displaystyle 3\sqrt{2} \approx 4.24	(3, 3)
\displaystyle \frac{\pi}{3}	3	\displaystyle \left(\frac{3}{2}, \frac{3\sqrt{3}}{2}\right)
\displaystyle \frac{\pi}{2}	0	(0, 0)
\displaystyle \frac{2\pi}{3}	-3	\displaystyle \left(\frac{3}{2}, -\frac{3\sqrt{3}}{2}\right)
\displaystyle \frac{3\pi}{4}	\displaystyle -3\sqrt{2} \approx -4.24	(3, -3)
\displaystyle \frac{5\pi}{6}	\displaystyle -3\sqrt{3} \approx -5.20	\displaystyle \left(\frac{9}{2}, -\frac{3\sqrt{3}}{2}\right)
\pi	-6	(6, 0)

From 0 to \displaystyle \frac{\pi}{2} the point moves from (6, 0) over the top to the pole. After \displaystyle \frac{\pi}{2}, r is negative, so each point is plotted backward, below the polar axis, and the point comes round the bottom back to (6, 0) at \theta = \pi. That is one full trip.

Figure 5Example 2. The points for \displaystyle 0 \le \theta \le \frac{\pi}{2} fill the upper half; the points with negative r fill the lower half.

To confirm, multiply by r: r^2 = 6r\cos\theta, so x^2 + y^2 = 6x and (x - 3)^2 + y^2 = 9. The graph is the circle with center (3, 0) and radius 3. The pole was on the original graph already (\displaystyle \theta = \frac{\pi}{2}), so multiplying by r added nothing.

Example 3

Converting r = -4 sin theta

Identify the graph of r = -4\sin\theta by converting, and check it with one point.

Solution

Multiply by r:

\begin{aligned} r^2 &= -4r\sin\theta \\ x^2 + y^2 + 4y &= 0 \\ x^2 + (y + 2)^2 &= 4 \end{aligned}

The graph is the circle with center (0, -2) and radius 2, below the pole; the pole was already on the graph, at \theta = 0. For 0 < \theta < \pi, \sin\theta > 0 makes r negative, so every point is plotted backward, below the polar axis, as the center says. Check \displaystyle \theta = \frac{\pi}{6}: \displaystyle r = -4 \cdot \frac{1}{2} = -2, the point \displaystyle x = -2\cos\frac{\pi}{6} = -\sqrt{3}, \displaystyle y = -2\sin\frac{\pi}{6} = -1, and \displaystyle (-\sqrt{3})^2 + (-1 + 2)^2 = 3 + 1 = 4.

Example 4

Testing r = 5 sin theta for symmetry

Erzhan tests r = 5\sin\theta for all three symmetries and uses the result to graph it from a short table.

Solution

Polar axis. Replace \theta by -\theta: r = 5\sin(-\theta) = -5\sin\theta, not the original. Replace (r, \theta) by (-r, \pi - \theta): -r = 5\sin(\pi - \theta) = 5\sin\theta, so r = -5\sin\theta. Both fail; no conclusion yet.

The line \displaystyle \theta = \frac{\pi}{2}. Replace \theta by \pi - \theta: r = 5\sin(\pi - \theta) = 5\sin\theta, the original. The graph is symmetric about the line \displaystyle \theta = \frac{\pi}{2}.

Pole. (-r, \theta) gives r = -5\sin\theta, and (r, \theta + \pi) gives r = 5\sin(\theta + \pi) = -5\sin\theta. Both fail.

By the symmetry, the half for \displaystyle 0 \le \theta \le \frac{\pi}{2} is enough; the other half is its mirror image across the vertical axis.

\theta	0	\displaystyle \frac{\pi}{6}	\displaystyle \frac{\pi}{4}	\displaystyle \frac{\pi}{3}	\displaystyle \frac{\pi}{2}
r	0	\displaystyle \frac{5}{2}	\displaystyle \frac{5\sqrt{2}}{2}	\displaystyle \frac{5\sqrt{3}}{2}	5

The points rise from the pole to \displaystyle (5, \frac{\pi}{2}), the top of the circle with center \displaystyle \left(0, \frac{5}{2}\right) and radius \displaystyle \frac{5}{2}. That circle really has no symmetry about the polar axis or the pole: \displaystyle (5, \frac{\pi}{2}) is on it, and its mirror image in both, the point (0, -5), lies 5 units below the pole, where the circle never goes.

Example 5

A test that fails while the symmetry holds

Gaukhar tests r = 4 for symmetry about the pole by replacing r with -r. She gets r = -4 and concludes the circle is not symmetric about the pole. What went wrong?

Solution

The equation r = -4 is not the equation r = 4, so the test fails. But the graph of r = 4 is the circle of radius 4 about the pole, and a half turn about its center maps it onto itself. The test missed it because each mirror image is on the graph under its other name: the image of (4, \theta) is (-4, \theta), which is also (4, \theta + \pi), and that name satisfies r = 4. The second test in the row, replacing \theta by \theta + \pi, gives r = 4 again and passes. A failed test is only a failed test; it never proves the symmetry is absent.

Example 6

Where two polar circles meet

Kairat finds where r = 6\sin\theta and \displaystyle r = 2\sqrt{3}\cos\theta intersect by setting the two right sides equal. Find his answer, and find the point he misses.

Solution

\displaystyle 6\sin\theta = 2\sqrt{3}\cos\theta gives \displaystyle \tan\theta = \frac{2\sqrt{3}}{6} = \frac{\sqrt{3}}{3}, so \displaystyle \theta = \frac{\pi}{6} in [0, \pi), where each circle is traced once. Then \displaystyle r = 6\sin\frac{\pi}{6} = 3, and \displaystyle 2\sqrt{3}\cos\frac{\pi}{6} = 3 as well. The point is \displaystyle P = \left(3, \frac{\pi}{6}\right), or \displaystyle \left(\frac{3\sqrt{3}}{2}, \frac{3}{2}\right).

Figure 6Example 6. The circles cross twice, but solving the two equations together finds only P.

The figure shows a second crossing at the pole. Both curves pass through it, but at different angles: r = 6\sin\theta is 0 at \theta = 0, and \displaystyle r = 2\sqrt{3}\cos\theta is 0 at \displaystyle \theta = \frac{\pi}{2}. No single \theta gives both, so setting the equations equal cannot find it. Rectangular form confirms both points: x^2 + y^2 = 6y and \displaystyle x^2 + y^2 = 2\sqrt{3}x give \displaystyle y = \frac{x}{\sqrt{3}}, and then \displaystyle \frac{4}{3}x^2 = 2\sqrt{3}x, so x = 0 or \displaystyle x = \frac{3\sqrt{3}}{2}. Always check the pole separately when intersecting polar curves.

Boundaries

These are the only circles of this simple kind. r = a\cos\theta and r = a\sin\theta give circles through the pole centered on an axis; r = a gives circles centered at the pole. A circle through the pole with center elsewhere is r = 2h\cos\theta + 2k\sin\theta for center (h, k), and a circle that misses the pole and is not centered there needs an equation with both r^2 and r in it, r^2 - 2r(h\cos\theta + k\sin\theta) + h^2 + k^2 = R^2 for radius R.

One trip needs an interval of length \pi. On [0, 2\pi) the circles r = a\cos\theta and r = a\sin\theta are traced twice. That matters when counting intersections or measuring how far a point travels along the curve.

The tests are one-way. A passed test proves the symmetry. A failed test proves nothing, even when both names fail; showing that a symmetry is absent needs a point of the graph whose mirror image is not on it.

Intersections can hide at the pole. Two polar curves can share the pole at different \theta, so solving their equations together misses it.

Where this goes next

The same tests and the same habit of tracing an interval show their full use on the curves in Limacons and cardioids (limaçons and cardioids, symmetric about one axis) and Rose curves, where the second name in a row is often the one that passes. All of them belong to Polar equations and their graphs.

Exercises

Graphing and converting

Give the rectangular form and name the graph; for a circle, give its center and radius.

r = 9
\displaystyle \theta = \frac{\pi}{6}
r = 11\sin\theta
r = -9\cos\theta
r = -3\sin\theta
For r = 7\cos\theta, find r at \displaystyle \theta = \frac{\pi}{3} and at \displaystyle \theta = \frac{2\pi}{3}, and give the rectangular coordinates of both points. How are the two points related?

Writing the equation

Write a polar equation for the circle with center (7, 0) and radius 7.
Write a polar equation for the circle with center (0, -1) and radius 1.
Write a polar equation for the circle whose diameter runs from the pole to the rectangular point (-5, 0).
The rectangular point (6, 6) is on r = 12\cos\theta. Find the \theta in [0, \pi) and the \theta in [\pi, 2\pi) that reach it, with the value of r for each.

Symmetry tests

Test each equation for symmetry about the polar axis, the line \displaystyle \theta = \frac{\pi}{2} and the pole, trying both names in each row. Then say which symmetries the graph actually has.

r = -6\sin\theta
r = 3\sec\theta
r = \theta, with \theta any real number
\displaystyle \theta = \frac{\pi}{3}
r = \cos\theta + \sin\theta

Reasoning

Explain why r = a\sin\theta traces its circle once as \theta runs over [0, \pi), and twice over [0, 2\pi).
For the rose r = \sin 2\theta (see Rose curves), show that replacing \theta by -\theta fails, yet the graph is symmetric about the polar axis.
Show that r = a\sin\theta is the circle with center \displaystyle \left(0, \frac{a}{2}\right) and radius \displaystyle \frac{|a|}{2}.
Explain why the graph of r = a\sin\theta is the graph of r = a\cos\theta turned \displaystyle \frac{\pi}{2} counterclockwise about the pole.
Kairat says r = 10\cos\theta and r = 10\sin\theta meet only at \displaystyle \left(5\sqrt{2}, \frac{\pi}{4}\right). Is he right?

Answers

x^2 + y^2 = 81, the circle with center (0, 0) and radius 9.
\displaystyle y = \frac{\sqrt{3}}{3}x, the line through the pole at angle \displaystyle \frac{\pi}{6}. Its slope is \displaystyle \tan\frac{\pi}{6}.
\displaystyle x^2 + \left(y - \frac{11}{2}\right)^2 = \frac{121}{4}, center \displaystyle \left(0, \frac{11}{2}\right), radius \displaystyle \frac{11}{2}. From x^2 + y^2 = 11y.
\displaystyle \left(x + \frac{9}{2}\right)^2 + y^2 = \frac{81}{4}, center \displaystyle \left(-\frac{9}{2}, 0\right), radius \displaystyle \frac{9}{2}. From x^2 + y^2 = -9x.
\displaystyle x^2 + \left(y + \frac{3}{2}\right)^2 = \frac{9}{4}, center \displaystyle \left(0, -\frac{3}{2}\right), radius \displaystyle \frac{3}{2}. From x^2 + y^2 = -3y.
\displaystyle r = \frac{7}{2} and \displaystyle r = -\frac{7}{2}. The points are \displaystyle \left(\frac{7}{4}, \frac{7\sqrt{3}}{4}\right) and \displaystyle \left(\frac{7}{4}, -\frac{7\sqrt{3}}{4}\right), mirror images across the polar axis, as the circle’s symmetry about that axis requires.
r = 14\cos\theta. The center is \displaystyle \left(\frac{a}{2}, 0\right) with \displaystyle \frac{a}{2} = 7.
r = -2\sin\theta. The center is \displaystyle \left(0, \frac{a}{2}\right) with \displaystyle \frac{a}{2} = -1.
r = -5\cos\theta. The center is \displaystyle \left(-\frac{5}{2}, 0\right), so a = -5; at \theta = 0, r = -5 reaches (-5, 0).
\displaystyle \theta = \frac{\pi}{4} with \displaystyle r = 6\sqrt{2}, and \displaystyle \theta = \frac{5\pi}{4} with \displaystyle r = -6\sqrt{2}. The point is at distance \displaystyle 6\sqrt{2} in direction \displaystyle \frac{\pi}{4}, and \displaystyle 12\cos\frac{5\pi}{4} = -6\sqrt{2} names the same point.
Polar axis: both names fail. Line \displaystyle \theta = \frac{\pi}{2}: (r, \pi - \theta) gives r = -6\sin\theta, pass. Pole: both fail. The graph, the circle with center (0, -3) and radius 3, is symmetric about the line \displaystyle \theta = \frac{\pi}{2} only.
Polar axis: \sec(-\theta) = \sec\theta, pass. Line \displaystyle \theta = \frac{\pi}{2} and pole: both names give r = -3\sec\theta, fail. The graph is the line x = 3, symmetric about the polar axis only.
Line \displaystyle \theta = \frac{\pi}{2}: (-r, -\theta) gives -r = -\theta, which is r = \theta, pass. The other five names give r = -\theta, r = \theta - \pi, r = \pi - \theta, r = -\theta and r = \theta + \pi, all fail. The graph is symmetric about the line \displaystyle \theta = \frac{\pi}{2} and about nothing else. To see that the other two symmetries are really absent, take the point \displaystyle \left(\frac{\pi}{2}, \frac{\pi}{2}\right) of the graph, which is \displaystyle \left(0, \frac{\pi}{2}\right) in rectangular form. Its mirror image in the polar axis and its image in the pole are the same point, \displaystyle \left(0, -\frac{\pi}{2}\right). Every polar name of that point is \displaystyle \left(\frac{\pi}{2}, \frac{3\pi}{2} + 2\pi k\right) or \displaystyle \left(-\frac{\pi}{2}, \frac{\pi}{2} + 2\pi k\right) for an integer k. To lie on r = \theta, a name needs its angle equal to its r: \displaystyle \frac{3\pi}{2} + 2\pi k = \frac{\pi}{2} needs 2\pi k = -\pi, and \displaystyle \frac{\pi}{2} + 2\pi k = -\frac{\pi}{2} also needs 2\pi k = -\pi. Neither has an integer k, so the image is not on the graph.
Polar axis: the names give \displaystyle \theta = -\frac{\pi}{3} and \displaystyle \theta = \frac{2\pi}{3}, fail; the line at \displaystyle -\frac{\pi}{3} is a different line, and indeed the graph is not symmetric. Line \displaystyle \theta = \frac{\pi}{2}: \displaystyle \theta = \frac{2\pi}{3} and \displaystyle \theta = -\frac{\pi}{3}, fail, and not symmetric. Pole: (-r, \theta) leaves \displaystyle \theta = \frac{\pi}{3}, pass, so the line is symmetric about the pole, although (r, \theta + \pi) gives \displaystyle \theta = -\frac{2\pi}{3}, a different equation for the same line.
Every name fails: they give r = \cos\theta - \sin\theta, r = \sin\theta - \cos\theta or r = -\cos\theta - \sin\theta. The graph, x^2 + y^2 = x + y, is the circle with center \displaystyle \left(\frac{1}{2}, \frac{1}{2}\right) and radius \displaystyle \frac{\sqrt{2}}{2}; it has none of the three symmetries, though it is symmetric about the line \displaystyle \theta = \frac{\pi}{4}.
On [0, \pi) each line through the pole is used once and meets the circle at one point besides the pole (the tangent line \theta = 0 only at the pole), so each point is reached once. Since a\sin(\theta + \pi) = -a\sin\theta, the point for \theta + \pi is (-r, \theta + \pi), the same point as for \theta, so [\pi, 2\pi) traces it again.
\sin(-2\theta) = -\sin 2\theta, so (r, -\theta) gives r = -\sin 2\theta: fail. But (-r, \pi - \theta) gives -r = \sin(2\pi - 2\theta) = -\sin 2\theta, so r = \sin 2\theta: pass, and the graph is symmetric about the polar axis.
r^2 = ar\sin\theta gives x^2 + y^2 = ay, and completing the square, \displaystyle x^2 + \left(y - \frac{a}{2}\right)^2 = \frac{a^2}{4}. The pole is on the original at \theta = 0, so nothing was added.
\displaystyle a\sin\theta = a\cos\left(\theta - \frac{\pi}{2}\right). If (r, \theta) is on r = a\cos\theta, then \displaystyle (r, \theta + \frac{\pi}{2}) satisfies r = a\sin\theta, and that point is (r, \theta) turned \displaystyle \frac{\pi}{2} about the pole.
No. Equating gives \tan\theta = 1 and the point \displaystyle \left(5\sqrt{2}, \frac{\pi}{4}\right), but both circles also pass through the pole, r = 10\cos\theta at \displaystyle \theta = \frac{\pi}{2} and r = 10\sin\theta at \theta = 0. They meet at two points.