Skip to main content
K12 LibreTexts

4.4.3: Systems of Polar Equations

  • Page ID
    14430
  • \( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \) \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)\(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\) \(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\)\(\newcommand{\AA}{\unicode[.8,0]{x212B}}\)

    Systems of Polar Equations

    You likely recall that when you graph multiple equations on the same line, you usually end up with locations on the graph where they intersect (unless you are graphing parallel lines!).

    The same is true when graphing equations in polar form, and/or on a polar graph. When you graph the intersection of multiple polar equations, you treat them just as you would rectangular equations, graph both and find the areas that are true for both equations.


    Systems of Polar Equations

    Polar equations can be graphed using polar coordinates. Graphing two polar equations on the same set of axes may result in having point(s) of intersection.

    All points on a polar graph are coordinates that make the equation valid. The coordinates of point(s) of intersection when substituted into each equation will make both of the equations valid.

    One method to find point(s) of intersection for two polar graphs is by setting the equations equal to each other.

    Call the first equation r1 and the second equation r2.

    Points of intersection are when r1 = r2, so set the equations equal and then solve the resulting trigonometric equation.


    Examples

    Example 1

    Find the intersection of \(\ r_{1}=3 \sin \theta\) and \(\ r_{2}=\sqrt{3} \cos \theta\)

    Solution

    Set the equations equal to each other: \(\ 3 \sin \theta=\sqrt{3} \cos \theta\)

    divide both sides of the equation by cos θ and 3: \(\ \frac{3 \sin \theta}{3 \cos \theta}=\frac{\sqrt{3} \cos \theta}{3 \cos \theta}\)

    Simplify: \(\ \frac{\sin \theta}{\cos \theta}=\frac{\sqrt{3}}{3}\)

    Use the identity: \(\ \frac{\sin \theta}{\cos \theta}=\tan \theta\)

    \(\ \tan \theta=\frac{\sqrt{3}}{3}\)

    \(\ \theta=\frac{\pi}{6} \text { or } \frac{7 \pi}{6}\)

    substitute \(\ \frac{\pi}{6}\) in either equation to obtain \(\ r=1.5\)

    substitute \(\ \frac{7 \pi}{6}\) in either equation to obtain -1.5

    NOTE: the coordinates \(\ \left(1.5, \frac{\pi}{6}\right)\) and \(\ \left(-1.5, \frac{7 \pi}{6}\right)\) represent the SAME polar point so there is only one solution to this equation.

    Are we done? If we look at the graphs of r1 and r2, we can see that there is another point of intersection:

    f-d_7b5bd83c9d2a50b9e9b9f4c0854ebb1c5f8af601fe7cc7f711b3c35d+IMAGE_TINY+IMAGE_TINY.jpg

    when θ = 0, r1 = 3 sin θ = 3 sin(0) = 0

    That means r1 = 3 sin θ goes through the pole (0, 0).

    For r2: when \(\ \theta=\frac{\pi}{2}\), r2 = 0 that is r2 = \(\ \sqrt{3} \cos \theta\) goes through the point \(\ \left(0, \frac{\pi}{2}\right)\).

    Therefore, both graphs go through the pole and the pole is a point of intersection.

    The pole was NOT revealed as a point of intersection using the first step! (Why? Hint: How many ways are there to represent the pole in polar coordinates?) This shows us that after you use algebraic methods to find intersections at points other than the pole, you should also check for intersections at the pole.

    Example 2

    Find the point(s) of intersection for the two graphs: \(\ r_{1}=1 \text { and } r_{2}=2 \sin 2 \theta\).

    Solution

    Set r1 = r2 and solve:

    \(\ \begin{array}{l}
    1=2 \sin 2 \theta \\
    \frac{1}{2}=\sin 2 \theta
    \end{array}\)

    Use a substitution \(\ \alpha=2 \theta\) to solve

    \(\ \begin{array}{l}
    \frac{1}{2}=\sin \alpha \\
    \alpha=\frac{\pi}{6}, \frac{5 \pi}{6}
    \end{array}\)

    Since \(\ \alpha=2 \theta\), solving for \(\ \theta\) gives us

    \(\ \theta=\frac{\pi}{12}, \frac{5 \pi}{12}\)

    But, recall that θ has the range 0 ≤ θ ≤ 2π. Since we solved with 0 ≤ α ≤ 2π we actually need to consider values of θ with 0 ≤ θ ≤ 4π. Why? Recall that sin(2θ) has two cycles between 0 and 2π, and so we add two more solutions,

    \(\ \alpha=\frac{13 \pi}{6}, \frac{17 \pi}{6}\)

    and since \(\ \alpha=2 \theta\),

    \(\ \theta=\frac{13 \pi}{12}, \frac{17 \pi}{12}\)

    Finally, we need to consider solutions when r = -1 because r = 1 and r = -1 are the same polar equation. So, solving

    \(\ -\frac{1}{2}=\sin \alpha\)

    \(\ \alpha=\frac{7 \pi}{6}, \frac{11 \pi}{6}\)

    Again, using α = 2θ and adding solutions for the repetition gives us four more solutions,

    \(\ \theta=\frac{7 \pi}{12}, \frac{11 \pi}{12}, \frac{19 \pi}{12}, \frac{21 \pi}{12}\)

    So in total, there are eight solutions to this set of equations.

    NOTE: Recall that solving trigonometric equations where the angle is θ requires looking at all potential values between 0 and 2π. When the angle is 2θ as it is in this case, be sure to look for all potential values between 0 and 4π. When the angle is 3θ as it is in this case, be sure to look for all potential values between 0 and 6π., and so on.

    Since r1 cannot equal 0, the pole is not on its graph and not a point of intersection.

    The graph reveals eight points of intersection which were found earlier.

    f-d_b321289defe2b07f906cd20ad092fb5a44d1de20e6952b4fb0f328fa+IMAGE_TINY+IMAGE_TINY.jpg

    Example 3

    Find point(s) of intersection, if any exist, for the following pair of equations: r1 = 2 and r2 = secθ.

    Solution

    Here we will use a table of values for each function, solving by quadrant. Recall that the period of sec θ is 2π.

    For the first quadrant:

    θ (angle) 0 π/6 π/4 π/3 π/2
    r1 (distance) 2 2 2 2 2
    r2 1 1.15 1.4 2 und

    For the second quadrant:

    θ (angle) 2π/3 3π/4 5π/6 π
    r1 (distance) 2 2 2 2
    r2 -2 -1.4 -1.15 -1

    For the third quadrant:

    θ (angle) 7π/6 5π/4 4π/3 3π/2
    r1 (distance) 2 2 2 2
    r2 -1.15 -1.4 -2 und

    For the fourth quadrant:

    θ (angle) 5π/3 7π/4 11π/6
    r1 (distance) 2 2 2 2
    r2 2 1.4 1.15 1

    f-d_44553349d9e3f11f86ce90169f3dfd637b0c008fd5e18098af957643+IMAGE_TINY+IMAGE_TINY.jpg

    Note that the 3rd and 4th quadrant repeat 1st and 2nd quadrant values.

    Observe in the table of values that (2, π/3) and (2, 5π/3) are the points of intersection. Look at the curves- the first equation yields a circle while the second yields a line. The maximum number of intersecting points for a line and a circle is 2. The two points have been found.

    Example 4

    Find the point(s) of intersection for this pair of Polar equations: r = 2 + 4 sinθ and θ = 60°.

    Solution

    The equation θ = 60o is a line making a 60° angle with the r axis.

    Make a table of values for r = 2 + 4 sin θ

    For the first quadrant:

    θ (angle) 0 30 45 60 90
    R (distance) 2 4 4.83 5.46 6

    For the second quadrant:

    θ (angle) 120 135 150 180
    R (distance) 5.46 4.83 4 2

    For the third quadrant:

    θ (angle) 210 225 240 270
    R (distance) 0 -.83 -1.46 -2

    For the fourth quadrant:

    θ (angle) 300 315 330 360
    R (distance) -1.46 -.83 0 2

    Notice there are two solutions in the table., (60, 5.46) and (240, -1.46) = (60, 1.46). Recall that when r < 0, you plot a point (r, θ), by rotating 180o (or π).

    Finally, we need to check the pole: r = 2 + 4 sin θ passes through the pole for θ = 330o, and θ = 60o also passes through the pole. Thus, the third point of intersection is (0, 0).

    f-d_9439515f18b15b17352371c5f1a9dabcefff656febc132683870a5fa+IMAGE_TINY+IMAGE_TINY.jpg

    Example 5

    Find the point(s) of intersection for this pair of Polar equations: r1 = 2 cosθ and r2 = 1.

    Solution

    Make a table:

    For the first quadrant:

    θ (angle) 0 π/6 π/4 π/3 π/2
    r1 (distance) 2 \(\ \sqrt{3}\) \(\ \sqrt{2}\) 1 0
    r2 1 1 1 1 1

    For the second quadrant:

    θ (angle) 2π/3 3π/4 5π/6 π
    r1 (distance) -1 \(\ -\sqrt{2}\) \(\ -\sqrt{3}\) -2
    r2 1 1 1 1

    For the third quadrant:

    θ (angle) 7π/6 5π/4 4π/3 3π/2
    r1 (distance) \(\ -\sqrt{3}\) \(\ -\sqrt{2}\) -1 0
    r2 1 1 1 1

    For the fourth quadrant:

    θ (angle) 5π/3 7π/4 11π/6
    r1 (distance) 1 \(\ \sqrt{2}\) \(\ \sqrt{3}\) 2
    r2 1 1 1 1

    So the unique solutions are at \(\ \theta=\frac{\pi}{3}, \frac{4 \pi}{3}\). There are also two repeated solutions in this set (can you find them?).

    Here is a graph showing the two solutions:

    f-d_dc3e20a2659fcd3bdee759089fa66e367bdb2ee247d1ca0d3d7bded6+IMAGE_TINY+IMAGE_TINY.jpg


    Review

    The graphs of r=1 and r=2cosθ are shown below.

    f-d_ed25de03539c422d440bd64c3389780bbf73efcd9347612407d56d29+IMAGE_TINY+IMAGE_TINY.png

    1. How many times do they intersect?
    2. In which quadrants do they intersect?
    3. At what points do the intersections occur?

    Based on the image below and the following information: the intersection of the graphs of r=cosθ and r=1−cosθ

    1. Identify how many times they intersect?
    2. At what points do the intersections occur?

    Find the points of intersection of the following pairs of curves.

    1. r=2 r=2cosθ
    2. r=sin2θ r=2sinθ
    3. r=2+2sinθ r=2−2cosθ
    4. r=3cosθ r=2−cosθ

    Find the point(s) of intersection for each system of equations. Graph to verify your solution.

    1. \(\ r_{1}=\csc \theta\ r_{2}=2 \sin \theta\)
    2. \(\ r_{1}=\cos \theta\ r_{2}=1+\sin \theta\)
    3. \(\ r_{1}=\sin \theta\ r_{2}=\sin 2 \theta\)
    4. \(\ r_{1}=-4 \sin \theta\ r_{2}=-4 \cos \theta\)
    5. \(\ r_{1}=1-2 \sin \theta\ r_{2}=\sqrt{9 \cos (\theta)}\)
    6. \(\ r_{1}=1-\cos \theta\ r_{2}=4 \cos (3 \theta)\)

    Review (Answers)

    To see the Review answers, open this PDF file and look for section 4.3.


    Vocabulary

    Term Definition
    Conic Conic sections are those curves that can be created by the intersection of a double cone and a plane. They include circles, ellipses, parabolas, and hyperbolas.
    Points of intersection Points of intersection are locations where two different equations have the same solutions.
    polar coordinates Polar coordinates describe locations on a grid using the polar coordinate system. The location of each point is determined by its distance from the pole and its angle with respect to the polar axis.
    pole The pole is the center point on a polar graph.
    quadrant A quadrant is one-fourth of the coordinate plane. The four quadrants are numbered using Roman Numerals I, II, III, and IV, starting in the top-right, and increasing counter-clockwise.

    This page titled 4.4.3: Systems of Polar Equations is shared under a CK-12 license and was authored, remixed, and/or curated by CK-12 Foundation via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.

    CK-12 Foundation
    LICENSED UNDER
    CK-12 Foundation is licensed under CK-12 Curriculum Materials License
    • Was this article helpful?