From Polar Equations to Familiar Shapes
Polar equations can describe the same lines and circles that appear in Cartesian graphing. Their equations may look unfamiliar, but the key is to translate \(r\cos\theta\) and \(r\sin\theta\) into the Cartesian coordinates \(x\) and \(y\). Once converted, completing the square or identifying a line equation reveals the shape and its location.
In a polar coordinate system, a point is described by a directed distance \(r\) from the pole and an angle \(\theta\). Negative values of \(r\) are allowed: \((r,\theta)\) and \((-r,\theta+\pi)\) represent the same point. The Cartesian conversion formulas are \(x=r\cos\theta\) and \(y=r\sin\theta\).
Polar Equations of Lines
A fixed angle \(\theta=\alpha\) gives the direction of a line through the pole. If \(r\) may be positive or negative, points in both directions along that line are included. If a graphing convention restricts \(r\) to nonnegative values, the same equation shows only a ray from the pole.
A line not passing through the pole can also be written in polar form. Start with a Cartesian line \(Ax+By=C\), where \(A\), \(B\), and \(C\) are constants and \(A\) and \(B\) are not both zero. Substituting \(x=r\cos\theta\) and \(y=r\sin\theta\) gives
When the factor \(A\cos\theta+B\sin\theta\) is nonzero, this can be solved for \(r\). For example, \(r\cos\theta=2\) is the vertical line \(x=2\). At angles where the factor is zero, there is no finite \(r\) satisfying the equation if \(C\ne 0\); this does not remove any points from the Cartesian line.
Worked Example: Sketch a Line Not Through the Pole
Graph \(r=\frac{4}{2\cos\theta+\sin\theta}\). Multiply both sides by the denominator:
Distribute \(r\), then use \(r\cos\theta=x\) and \(r\sin\theta=y\):
The \(x\)-intercept is found by setting \(y=0\): \(2x=4\), so \(x=2\). The \(y\)-intercept is found by setting \(x=0\): \(y=4\). Sketch the straight line through \((2,0)\) and \((0,4)\). Check: at \((2,0)\), \(\theta=0\) and \(r=2\), so the original right-hand expression is \(4/(2\cdot1+0)=2\). At \((0,4)\), \(\theta=\pi/2\) and \(r=4\), giving \(4/(0+1)=4\). Both points satisfy the polar equation.
Circles of the Form \(r=a\cos\theta\) or \(r=a\sin\theta\)
These equations describe circles that pass through the pole. Convert the cosine form first:
Completing the square in \(x\) identifies the center and radius:
Therefore, for \(a\ne0\), the circle has center \((a/2,0)\) and radius \(|a|/2\). The absolute value matters because a radius is nonnegative, even if \(a\) is negative. If \(a=0\), the equation describes only the pole.
Similarly, starting with \(r=a\sin\theta\) gives \(x^2+y^2=ay\). Completing the square in \(y\) gives
Thus this is a circle centered at \((0,a/2)\) with radius \(|a|/2\), or just the pole if \(a=0\). The sign of \(a\) determines whether the center lies above or below the pole for the sine form, and to the right or left for the cosine form.
Worked Example: Sketch \(r=6\cos\theta\)
Use \(r^2=x^2+y^2\) and \(r\cos\theta=x\):
Complete the square:
The graph is a circle centered at \((3,0)\) with radius \(3\). It passes through the pole and has its rightmost point at \((6,0)\). Check the conversion with the point at \(\theta=\pi/4\): \(r=6(\sqrt{2}/2)=3\sqrt{2}\), so \(x=r\cos\theta=3\) and \(y=r\sin\theta=3\). The Cartesian equation gives \((3-3)^2+3^2=9\), as required.
Worked Example: Sketch \(r=4\sin\theta\)
Use \(r\sin\theta=y\) to convert the equation:
Complete the square in \(y\):
The graph is a circle centered at \((0,2)\) with radius \(2\). Its lowest point is the pole and its highest point is \((0,4)\). Check the highest point in polar form: at \(\theta=\pi/2\), \(r=4\), so \(x=4\cos(\pi/2)=0\) and \(y=4\sin(\pi/2)=4\). Substitution gives \(0^2+(4-2)^2=4\), confirming that the point lies on the circle.
Constant-Radius Circles
An equation \(r=c\), where \(c\) is constant, fixes the distance from every point on the graph to the pole. As \(\theta\) varies through a full turn, the graph is a circle centered at the pole. Its radius is \(|c|\), since a negative polar radius represents the same distance in the opposite direction.
This conclusion can also be checked in Cartesian coordinates. Squaring \(r=c\) gives \(r^2=c^2\), and \(r^2=x^2+y^2\). The equation is therefore \(x^2+y^2=c^2\), a circle centered at the origin with radius \(|c|\). For \(c=0\), the graph is just the pole.
Worked Example: Sketch \(r=3\)
The radius is fixed at \(3\), so the graph is a circle centered at the pole with radius \(3\). In Cartesian form,
The circle passes through \((3,0)\), \((0,3)\), \((-3,0)\), and \((0,-3)\). Check, for example, the point \((0,-3)\): its distance from the pole is \(\sqrt{0^2+(-3)^2}=3\), so it satisfies the constant-radius equation. In polar coordinates it can be represented by \((r,\theta)=(3,3\pi/2)\).
A Reliable Sketching Method
A quick sketch should show the shape, its position relative to the pole, and enough points to confirm the scale. Converting to Cartesian form is usually the most dependable way to determine the center, radius, or intercepts. The following sequence applies to these common lines and circles.
A fixed angle describes a line through the pole; a constant \(r\) describes a circle centered at the pole. Equations involving \(r\cos\theta\) or \(r\sin\theta\) may describe shifted circles or lines.
Use \(x=r\cos\theta\), \(y=r\sin\theta\), and \(r^2=x^2+y^2\). Rearrange the result into a familiar line or circle equation.
For a circle, mark its center and radius. For a line, find convenient intercepts or use its slope and intercept form.
Substitute a convenient point into the Cartesian equation and, where useful, verify its polar representation. This can catch sign or location errors before sketching.
Why the Conversion Works—and a Common Pitfall
The conversion is not a separate trick for each equation. It follows directly from the coordinate definition: a point at directed distance \(r\) and angle \(\theta\) has horizontal coordinate \(r\cos\theta\) and vertical coordinate \(r\sin\theta\). Squaring and adding these coordinates gives \(x^2+y^2=r^2(\cos^2\theta+\sin^2\theta)=r^2\). These identities turn polar equations into Cartesian equations whose graphs can be recognized.
A common mistake is to treat the number multiplying \(\cos\theta\) as the radius of a circle centered at the pole. For \(r=6\cos\theta\), that incorrect reading predicts a circle of radius \(6\) centered at the pole. The conversion shows why that is wrong:
The actual center is \((3,0)\) and the radius is \(3\). In contrast, \(r=6\) is the constant-radius equation that gives a circle centered at the pole with radius \(6\). The two equations may look similar, but the trigonometric factor changes the circle's location and size.
Negative \(r\) values also deserve attention. They do not mean that a point has a negative distance from the pole; they mean that the point is plotted in the direction opposite to the angle \(\theta\). When graphing the equations in this tutorial, rely on the Cartesian equation to identify the complete locus, rather than assuming that every point must be represented with \(r\ge0\).
Check Your Understanding
Use a Cartesian conversion or a geometric feature to justify each answer.
- What line does \(r\cos\theta=5\) represent?
- Find the center and radius of the circle \(r=8\sin\theta\).
- What is the radius and center of the graph \(r=-4\)?
- Convert \(r=10\cos\theta\) into a Cartesian circle equation and identify its center.
- Why does \(\theta=\pi/3\) describe a line only when both positive and negative \(r\) values are allowed?
