What happens to the circle when its distance from the pole is pushed out on one side and pulled in on the other by , and why does the result sometimes have a dent, a point, or a loop inside a loop?

Precalculus

Limacons and cardioids

What happens to the circle r = a when its distance from the pole is pushed out on one side and pulled in on the other by b\cos\theta, and why does the result sometimes have a dent, a point, or a loop inside a loop?

You need

Plotting polar pointsPrecalculus

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

Why this exists

The circles of Polar circles and symmetry tests come from r = a, a fixed distance, and r = b\cos\theta, a distance swinging between b and -b. What curve does their sum trace?

The answer is a family of curves called limaçons. They are easy in polar form, awkward in rectangular form, and show plainly what a negative r does (Plotting polar points).

Definition

A limaçon is the graph of a polar equation

r = a + b\cos\theta \qquad \text{or} \qquad r = a + b\sin\theta

with a and b nonzero real numbers. A cardioid is the limaçon with |a| = |b|; its name means heart shaped.

Its shape is named by the ratio \displaystyle \frac{|a|}{|b|}:

Ratio	Shape
\displaystyle 0 < \frac{\vert a \vert}{\vert b \vert} < 1	limaçon with an inner loop
\displaystyle \frac{\vert a \vert}{\vert b \vert} = 1	cardioid
\displaystyle 1 < \frac{\vert a \vert}{\vert b \vert} < 2	dimpled limaçon (one loop with a dent)
\displaystyle \frac{\vert a \vert}{\vert b \vert} \ge 2	convex limaçon (one loop, no dent)

Intuition

Think of r = a + b\cos\theta, with a, b > 0, as the circle r = a bent: the distance grows to a + b at \theta = 0, shrinks to a - b at \theta = \pi, and stays a straight up and down.

When a is much larger than b, the circle becomes an egg, wider on the right.
As a shrinks toward b, the left side flattens and then dents inward.
At a = b the left side reaches the pole (r = 0 at \theta = \pi) and the dent closes into a point, the cusp of the cardioid.
When a < b, r near \theta = \pi is negative. Plotted backward, those points land right of the pole, inside the big loop: the inner loop.

Figure 1The same varying term 2\cos\theta with four constants. The ratio \displaystyle \frac{a}{|b|} alone decides the shape.

Results

Symmetry. The graph of r = a + b\cos\theta is symmetric about the polar axis, since replacing \theta by -\theta gives back the same equation. The graph of r = a + b\sin\theta is symmetric about the line \displaystyle \theta = \frac{\pi}{2}, since replacing \theta by \pi - \theta does. These are the symmetry tests of Polar circles and symmetry tests.

The constant may be taken positive. r = a + b\cos\theta and r = -a + b\cos\theta have the same graph (likewise with \sin\theta). So from here on a > 0.

The pole. The graph passes through the pole exactly when a + b\cos\theta = 0 has a solution, that is, when \displaystyle \cos\theta = -\frac{a}{b} has one: when a \le |b|. A cardioid reaches the pole at one angle in [0, 2\pi); a limaçon with an inner loop reaches it at two.

Largest and smallest distance. The largest distance from the pole is a + |b|, reached in the direction where b\cos\theta = |b| (for the sine form, b\sin\theta = |b|). When a > |b| the smallest distance is a - |b|, in the opposite direction.

The inner loop. When a < |b|, the inner loop is traced exactly for the \theta with r < 0. Its far point is at distance |b| - a from the pole, on the same side as the far point of the big loop.

Why it is true

The constant may be taken positive. The point (a + b\cos\theta, \theta) is also named (-a - b\cos\theta, \theta + \pi), because (r, \theta) and (-r, \theta + \pi) are the same point. Since \cos(\theta + \pi) = -\cos\theta, that second name is (-a + b\cos(\theta + \pi), \theta + \pi), which satisfies r = -a + b\cos\theta. So every point of one graph is on the other, and the argument runs both ways. With \sin(\theta + \pi) = -\sin\theta the same steps work for the sine form.

The pole, the largest distance, the inner loop. As b\cos\theta fills [-|b|, |b|], r fills [a - |b|, a + |b|], so it reaches 0 only when a \le |b|. When a < |b| the most negative r, a - |b|, is plotted backward: distance |b| - a on the side of the big loop.

Where the dent begins. Take r = a + b\cos\theta with 0 < b < a, and write c = \cos\theta. The x-coordinate of the point is

x = r\cos\theta = ac + bc^2 = b\left(c + \frac{a}{2b}\right)^2 - \frac{a^2}{4b}

a parabola in c with its lowest value at \displaystyle c = -\frac{a}{2b}. The point on the polar axis line, at \theta = \pi, has c = -1 and x = b - a.

If a < 2b, then \displaystyle -\frac{a}{2b} lies inside (-1, 1), and two points of the curve, above and below the axis, reach \displaystyle x = -\frac{a^2}{4b}. That is farther left than b - a, because \displaystyle (b - a) - \left(-\frac{a^2}{4b}\right) = \frac{(a - 2b)^2}{4b} > 0. The curve goes out past the axis point on both sides and comes back in to it: a dent.
If a \ge 2b, then \displaystyle -\frac{a}{2b} \le -1, so x increases as c runs from -1 to 1, and the only leftmost point is the one on the axis. No dent forms there. That the whole curve is then convex is also true, but proving it needs a way to measure how a curve bends, which comes with calculus.

Worked examples

Example 1

Naming four limaçons

For each equation, name the shape, the line of symmetry, and the largest value of |r| with the point where it occurs: r = 6 - 6\cos\theta, r = 3 - 8\sin\theta, r = 7 + 4\cos\theta and r = 6 - 3\sin\theta.

Solution

The shape comes from the ratio, the symmetry from the function, and the largest |r| is |a| + |b|, where the varying term has the sign of a and its largest size.

r = 6 - 6\cos\theta: ratio 1, a cardioid, symmetric about the polar axis; |r| = 12 at \theta = \pi, the point (-12, 0).
r = 3 - 8\sin\theta: ratio \displaystyle \frac{3}{8}, an inner loop, symmetric about the line \displaystyle \theta = \frac{\pi}{2}; |r| = 11 at \displaystyle \theta = \frac{3\pi}{2}, the point (0, -11).
r = 7 + 4\cos\theta: ratio \displaystyle \frac{7}{4}, dimpled, symmetric about the polar axis; r = 11 at \theta = 0, the point (11, 0).
r = 6 - 3\sin\theta: ratio 2, convex, symmetric about the line \displaystyle \theta = \frac{\pi}{2}; r = 9 at \displaystyle \theta = \frac{3\pi}{2}, the point (0, -9).

The cardioid reaches the pole at \theta = 0; the inner loop limaçon where \displaystyle \sin\theta = \frac{3}{8}, at \theta \approx 0.38 and \theta \approx 2.76; the other two never, since r \ge 3.

Example 2

A cardioid from a table

Graph r = 3 + 3\sin\theta by a table and its symmetry.

Solution

The ratio is 1, so this is a cardioid. Replacing \theta by \pi - \theta gives r = 3 + 3\sin(\pi - \theta) = 3 + 3\sin\theta, so the graph is symmetric about the line \displaystyle \theta = \frac{\pi}{2}, and the right half, \displaystyle -\frac{\pi}{2} \le \theta \le \frac{\pi}{2}, is enough to compute.

\theta	r = 3 + 3\sin\theta
\displaystyle -\frac{\pi}{2}	0
\displaystyle -\frac{\pi}{3}	\displaystyle 3 - \frac{3\sqrt{3}}{2} \approx 0.40
\displaystyle -\frac{\pi}{6}	\displaystyle \frac{3}{2}
0	3
\displaystyle \frac{\pi}{6}	\displaystyle \frac{9}{2}
\displaystyle \frac{\pi}{3}	\displaystyle 3 + \frac{3\sqrt{3}}{2} \approx 5.60
\displaystyle \frac{\pi}{2}	6

The point starts at the pole, swings out through (3, 0) and rises to the top, \displaystyle (6, \frac{\pi}{2}), the rectangular point (0, 6) and the largest distance. Reflecting gives the left half. The pole is reached only at \displaystyle \theta = \frac{3\pi}{2}, where the two halves meet in a sharp point, the cusp, at the bottom of the heart.

Figure 2Example 2. The table fills the right half; the mirror in the line \displaystyle \theta = \frac{\pi}{2} fills the left.

Example 3

Tracing an inner loop

Graph r = 1 + 2\cos\theta, find where it passes through the pole, and say which values of \theta trace the inner loop.

Solution

The ratio is \displaystyle \frac{1}{2}, so there is an inner loop. The pole: 1 + 2\cos\theta = 0 gives \displaystyle \cos\theta = -\frac{1}{2}, so \displaystyle \theta = \frac{2\pi}{3} or \displaystyle \theta = \frac{4\pi}{3}. The largest r is 1 + 2 = 3, at \theta = 0. The graph is symmetric about the polar axis, so a table on [0, \pi] is enough; the last column is (r\cos\theta, r\sin\theta).

\theta	r	(x, y)
0	3	(3, 0)
\displaystyle \frac{\pi}{3}	2	\displaystyle (1, \sqrt{3})
\displaystyle \frac{\pi}{2}	1	(0, 1)
\displaystyle \frac{2\pi}{3}	0	(0, 0)
\displaystyle \frac{3\pi}{4}	\displaystyle 1 - \sqrt{2} \approx -0.41	\displaystyle \left(\frac{2 - \sqrt{2}}{2}, \frac{\sqrt{2} - 2}{2}\right)
\displaystyle \frac{5\pi}{6}	\displaystyle 1 - \sqrt{3} \approx -0.73	\displaystyle \left(\frac{3 - \sqrt{3}}{2}, \frac{1 - \sqrt{3}}{2}\right)
\pi	-1	(1, 0)

From 0 to \displaystyle \frac{2\pi}{3} the point runs from (3, 0) over the top to the pole. After that r is negative, so the directions are in quadrant II but the points land in quadrant IV, ending at (1, 0) when \theta = \pi. The mirror image across the polar axis completes both loops. The inner loop is traced for \displaystyle \frac{2\pi}{3} < \theta < \frac{4\pi}{3}, exactly where r < 0, and its far point (1, 0) is on the same side as the big loop’s far point (3, 0).

Figure 3Example 3. The points with r < 0, plotted backward, make the small loop, on the same side of the pole as the large one.

Example 4

Finding the dent

Leszek is told that r = 5 + 4\cos\theta is dimpled. Find the leftmost points of the curve and show where the dent is.

Solution

The ratio is \displaystyle \frac{5}{4}, between 1 and 2. With c = \cos\theta,

x = r\cos\theta = 5c + 4c^2 = 4\left(c + \frac{5}{8}\right)^2 - \frac{25}{16}

so the smallest x is \displaystyle -\frac{25}{16}, at \displaystyle \cos\theta = -\frac{5}{8}. There \displaystyle r = 5 + 4\left(-\frac{5}{8}\right) = \frac{5}{2} and \displaystyle \sin\theta = \pm\sqrt{1 - \frac{25}{64}} = \pm\frac{\sqrt{39}}{8}, so the leftmost points are \displaystyle \left(-\frac{25}{16}, \pm\frac{5\sqrt{39}}{16}\right), one above and one below the axis.

The point on the axis, at \theta = \pi, has r = 1: the rectangular point (-1, 0), to the right of the leftmost points. Between them the curve swings back in to (-1, 0): that is the dent. With constant 8 instead, x = 8c + 4c^2 = 4(c + 1)^2 - 4 is smallest only at c = -1: no dent.

Figure 4Example 4. The leftmost points are off the axis, so the curve bends inward at (-1, 0).

Example 5

An equation from two points

Kazimierz knows a limaçon r = a + b\sin\theta with a > 0 and b > 0 has an inner loop. Its big loop reaches up to the point (0, 7) and its inner loop reaches up to (0, 3). Find a and b, and the angles in [0, 2\pi) where the curve passes through the pole.

Solution

The largest r is a + b, at \displaystyle \theta = \frac{\pi}{2}, the point (0, a + b); so a + b = 7. The inner loop’s far point comes from the most negative r, which is a - b at \displaystyle \theta = \frac{3\pi}{2}. Plotted backward, it lands at distance b - a in the direction \displaystyle \frac{\pi}{2}, the point (0, b - a); so b - a = 3. Adding the two equations, 2b = 10, so b = 5 and a = 2:

r = 2 + 5\sin\theta

The ratio \displaystyle \frac{2}{5} is below 1, as an inner loop needs. The pole: \displaystyle \sin\theta = -\frac{2}{5}, which gives (inverse sine) \displaystyle \theta = \pi + \arcsin\frac{2}{5} \approx 3.55 and \displaystyle \theta = 2\pi - \arcsin\frac{2}{5} \approx 5.87.

Example 6

A negative constant

Chenoa graphs r = -3 + 3\sin\theta. Every value of r is zero or negative. What is the graph?

Solution

By the rule for the constant, r = -3 + 3\sin\theta has the same graph as r = 3 + 3\sin\theta, the cardioid of Example 2. A single point confirms it. At \displaystyle \theta = \frac{\pi}{6}, \displaystyle r = -3 + \frac{3}{2} = -\frac{3}{2}: the point \displaystyle \frac{3}{2} units out in the direction \displaystyle \frac{7\pi}{6}. On the cardioid of Example 2, \displaystyle \theta = \frac{7\pi}{6} gives \displaystyle r = 3 - \frac{3}{2} = \frac{3}{2}, the same point. The top, too: \displaystyle \theta = \frac{3\pi}{2} gives r = -6, which is 6 units in the direction \displaystyle \frac{\pi}{2}, the point (0, 6). The two equations reach each point at different angles, but the set of points is the same.

Boundaries

Both constants must be nonzero. With a = 0, r = b\cos\theta is a circle through the pole, traced twice over [0, 2\pi) (see Polar circles and symmetry tests). With b = 0, r = a is a circle about the pole.

The two boundary ratios are exact. Ratio 1 is the cardioid, a point at the pole with no inner loop. Ratio 2 is convex: the curve is flattest at the point opposite its far point but does not dent.

The angle must be plain \theta. r = a + b\cos 2\theta is not a limaçon. A shifted angle, r = a + b\cos(\theta - \alpha), is a limaçon turned by \alpha about the pole.

Where this goes next

In the Rose curves the angle inside the cosine is multiplied, and the same habits apply: find where r = 0, where |r| is largest, and where r is negative. All belong to Polar equations and their graphs.

Exercises

Naming the shape

Name the shape and the line of symmetry, and give the largest value of |r| with the rectangular point where it occurs.

r = 4 + 4\cos\theta
r = 3 - 7\sin\theta
r = 9 + 5\cos\theta
r = 11 - 3\sin\theta
r = 7 - 5\cos\theta
r = -2 + 6\sin\theta. Also give an equation with a positive constant for the same graph.

Key points

For r = 2 + 2\sin\theta, find where it reaches the pole in [0, 2\pi), its largest r, and where it crosses the polar axis line.
For \displaystyle r = \sqrt{2} + 2\cos\theta, find the angles in [0, 2\pi) where r = 0, the interval of \theta that traces the inner loop, and the inner loop’s far point.
For r = 4 - 8\sin\theta, find where r = 0 in [0, 2\pi), the largest |r|, and the far point of the inner loop.
For r = 2 + 3\cos\theta, find to two decimal places the angles in [0, 2\pi) where r = 0, and the far point of the inner loop.
For r = 9 + 4\sin\theta, find the largest and smallest distance from the pole and where each occurs. Name the shape.

Graphing by a table

Make a table for r = 2 - 2\cos\theta at \displaystyle \theta = 0, \frac{\pi}{3}, \frac{\pi}{2}, \frac{2\pi}{3}, \pi, give each point in rectangular form, and use symmetry to describe the whole graph.
Make a table for r = 2 - 4\sin\theta at \displaystyle \theta = 0, \frac{\pi}{6}, \frac{\pi}{2}, \frac{5\pi}{6}, \pi, \frac{7\pi}{6}, \frac{3\pi}{2}, \frac{11\pi}{6}, and say which part of the graph is the inner loop.
Find the leftmost points of r = 4 + 3\cos\theta and compare them with the point at \theta = \pi.

Writing the equation

A cardioid r = a + b\cos\theta with a > 0 passes through the pole at \theta = 0 and reaches farthest at (-10, 0). Write its equation.
A limaçon r = a + b\cos\theta with a > 0, b > 0 crosses the polar axis line at (7, 0) and (-3, 0). Find its equation and name its shape.
A limaçon r = a - b\sin\theta with a > 0, b > 0 has an inner loop. Its big loop reaches down to (0, -9) and its inner loop down to (0, -1). Find its equation.

Reasoning

Explain why r = a + b\cos\theta, with a > 0 and b > 0, passes through the pole exactly when a \le b, and how many times in [0, 2\pi).
Show that the graph of r = a + b\sin\theta is the graph of r = a + b\cos\theta turned \displaystyle \frac{\pi}{2} counterclockwise about the pole.
Show that the graph of r = a - b\cos\theta is the mirror image of the graph of r = a + b\cos\theta across the line \displaystyle \theta = \frac{\pi}{2}.
Leszek says r = 6 + 3\sin\theta has a dent because its ratio is not above 2. Is he right?
Show that every point of r = 3 + 3\cos\theta satisfies (x^2 + y^2 - 3x)^2 = 9(x^2 + y^2).

Answers

Cardioid (ratio 1), symmetric about the polar axis; largest r = 8 at \theta = 0, the point (8, 0).
Inner loop (ratio \displaystyle \frac{3}{7}), symmetric about the line \displaystyle \theta = \frac{\pi}{2}; largest |r| = 10 at \displaystyle \theta = \frac{3\pi}{2}, the point (0, -10).
Dimpled (ratio \displaystyle \frac{9}{5}), symmetric about the polar axis; largest r = 14 at \theta = 0, the point (14, 0).
Convex (ratio \displaystyle \frac{11}{3} \ge 2), symmetric about the line \displaystyle \theta = \frac{\pi}{2}; largest r = 14 at \displaystyle \theta = \frac{3\pi}{2}, the point (0, -14).
Dimpled (ratio \displaystyle \frac{7}{5}), symmetric about the polar axis; largest r = 12 at \theta = \pi, the point (-12, 0).
Inner loop (ratio \displaystyle \frac{1}{3}), symmetric about the line \displaystyle \theta = \frac{\pi}{2}; same graph as r = 2 + 6\sin\theta. Largest |r| = 8: r = -8 at \displaystyle \theta = \frac{3\pi}{2}, the point (0, 8).
Pole at \displaystyle \theta = \frac{3\pi}{2}. Largest r = 4 at \displaystyle \theta = \frac{\pi}{2}, the point (0, 4). At \theta = 0 and \theta = \pi, r = 2: the points (2, 0) and (-2, 0), and the line also meets the curve at the pole.
\displaystyle \cos\theta = -\frac{\sqrt{2}}{2} gives \displaystyle \theta = \frac{3\pi}{4} and \displaystyle \frac{5\pi}{4}. The inner loop is \displaystyle \frac{3\pi}{4} < \theta < \frac{5\pi}{4}, where r < 0. Far point: \displaystyle r = \sqrt{2} - 2 at \theta = \pi, plotted at \displaystyle (2 - \sqrt{2}, 0).
\displaystyle \sin\theta = \frac{1}{2} gives \displaystyle \theta = \frac{\pi}{6} and \displaystyle \frac{5\pi}{6}. Largest |r| = 12, r = 12 at \displaystyle \theta = \frac{3\pi}{2}, the point (0, -12). Inner loop far point: r = -4 at \displaystyle \theta = \frac{\pi}{2}, plotted at (0, -4).
\displaystyle \cos\theta = -\frac{2}{3}: \displaystyle \theta = \arccos\left(-\frac{2}{3}\right) \approx 2.30 and \displaystyle 2\pi - \arccos\left(-\frac{2}{3}\right) \approx 3.98. Far point: r = -1 at \theta = \pi, plotted at (1, 0).
Largest 13 at \displaystyle \theta = \frac{\pi}{2}, the point (0, 13); smallest 5 at \displaystyle \theta = \frac{3\pi}{2}, the point (0, -5). Convex, ratio \displaystyle \frac{9}{4} \ge 2; never at the pole.
r = 0, 1, 2, 3, 4; points (0, 0), \displaystyle \left(\frac{1}{2}, \frac{\sqrt{3}}{2}\right), (0, 2), \displaystyle \left(-\frac{3}{2}, \frac{3\sqrt{3}}{2}\right), (-4, 0). Mirror in the polar axis for the lower half: a cardioid with its cusp at the pole and its far point at (-4, 0).
r = 2, 0, -2, 0, 2, 4, 6, 4. The inner loop is \displaystyle \frac{\pi}{6} < \theta < \frac{5\pi}{6}, where r < 0, reaching (0, -2); the big loop reaches (0, -6).
\displaystyle x = 4c + 3c^2 = 3\left(c + \frac{2}{3}\right)^2 - \frac{4}{3}, smallest at \displaystyle \cos\theta = -\frac{2}{3}, r = 2: the points \displaystyle \left(-\frac{4}{3}, \pm\frac{2\sqrt{5}}{3}\right). The point (-1, 0) at \theta = \pi is to their right: a dent.
r = 5 - 5\cos\theta. r = 0 at \theta = 0 needs b = -a; then r(\pi) = 2a = 10.
r = 5 + 2\cos\theta, convex. (-3, 0) needs r(\pi) = a - b = 3, and a + b = 7.
r = 4 - 5\sin\theta. Largest r = a + b = 9 at \displaystyle \theta = \frac{3\pi}{2}; the most negative, a - b at \displaystyle \theta = \frac{\pi}{2}, plots at (0, a - b), so b - a = 1.
r = 0 needs \displaystyle \cos\theta = -\frac{a}{b}, possible only when \displaystyle \frac{a}{b} \le 1. When a = b, \cos\theta = -1 has one solution, \theta = \pi; when a < b, \displaystyle -1 < -\frac{a}{b} < 0 gives two.
\displaystyle \sin\theta = \cos\left(\theta - \frac{\pi}{2}\right), so if (r, \theta) is on r = a + b\cos\theta, then \displaystyle \left(r, \theta + \frac{\pi}{2}\right) satisfies r = a + b\sin\theta; that point is (r, \theta) turned \displaystyle \frac{\pi}{2} about the pole. Turning back by \displaystyle \frac{\pi}{2} runs the same steps in reverse, so the two graphs match exactly.
The mirror image of (r, \theta) across that line is (r, \pi - \theta), and a + b\cos(\pi - \theta) = a - b\cos\theta. So a point is on one graph exactly when its image is on the other.
No. The ratio is exactly 2, which is convex. The lowest points: y = 6s + 3s^2 = 3(s + 1)^2 - 3 with s = \sin\theta is smallest only at s = -1, the point (0, -3), so there is no dent.
Multiply by r: r^2 = 3r + 3r\cos\theta, so x^2 + y^2 - 3x = 3r. Squaring gives (x^2 + y^2 - 3x)^2 = 9r^2 = 9(x^2 + y^2).