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# Systems of Polar Equations

## Points of intersection or locations where two equations have the same solution.

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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 \begin{align*}r_1 = 3 \ \mbox{sin} \theta \end{align*} and \begin{align*} r_2=\sqrt{3} \mbox{ cos}\ \theta \end{align*}.

Set the equations equal to each other: \begin{align*}3\ \mbox{sin}\ \theta =\sqrt{3}\ \mbox{cos}\ \theta \end{align*}

divide both sides of the equation by cos θ and 3: \begin{align*}\frac{3\ \mbox{sin}\ \theta}{3\ \mbox{cos}\ \theta} = \frac{\sqrt{3}\ \mbox{cos}\ \theta}{3\ \mbox{cos}\ \theta}\end{align*}

Simplify: \begin{align*}\frac{\mbox{sin}\ \theta}{\mbox{cos}\ \theta} = \frac{\sqrt{3}}{3}\end{align*}

Use the identity: \begin{align*}\frac{\mbox{sin}\ \theta}{\mbox{cos}\ \theta} = \mbox{tan}\ \theta\end{align*}

\begin{align*}\mbox{tan}\ \theta = \frac{\sqrt{3}}{3}\end{align*}

\begin{align*}\theta = \frac{\pi}{6}\end{align*} or \begin{align*}\frac{7 \pi}{6}\end{align*}

substitute \begin{align*}\frac{\pi}{6}\end{align*} in either equation to obtain r = 1.5

substitute \begin{align*}\frac{7\pi}{6}\end{align*} in either equation to obtain -1.5

NOTE: the coordinates \begin{align*}\left (1.5,\frac{\pi}{6}\right )\end{align*} and \begin{align*}\left (-1.5,\frac{7\pi}{6}\right)\end{align*} 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:

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 \begin{align*} \theta = \frac{\pi}{2}\end{align*}, r2 = 0 that is \begin{align*}r_2 = \sqrt{3}\ \mbox{cos}\ \theta\end{align*} goes through the point \begin{align*}\left (0,\frac{\pi}{2}\right )\end{align*}.

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: r1 = 1 and r2 = 2 sin 2θ.

Set r1 = r2 and solve:

\begin{align*}1 = 2\ \mbox{sin}\ 2\theta\end{align*}

\begin{align*}\frac{1}{2} = \mbox{sin}\ 2\theta\end{align*}

Use a substitution α = 2θ to solve

\begin{align*}\frac{1}{2} = \mbox{sin}\ \alpha \end{align*}

\begin{align*}\alpha = \frac{\pi}{6}, \frac{5\pi}{6}\end{align*}

Since α = 2θ , solving for θ gives us

\begin{align*}\theta = \frac{\pi}{12}, \frac{5\pi}{12}\end{align*}

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,

\begin{align*}\alpha = \frac{13\pi}{6}, \frac{17\pi}{6}\end{align*}

and since α = 2θ,

\begin{align*}\theta = \frac{13\pi}{12}, \frac{17\pi}{12}\end{align*}

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

\begin{align*}-\frac{1}{2} = \mbox{sin}\ \alpha\end{align*}

\begin{align*}\alpha = \frac{7\pi}{6}, \frac{11\pi}{6}\end{align*}

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

\begin{align*}\theta = \frac{7\pi}{12}, \frac{11\pi}{12}, \frac{19\pi}{12}, \frac{21\pi}{12}\end{align*}

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.

#### Example 3

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

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

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

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

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

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

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 equationsr = 2 + 4 sinθ and θ = 60°.

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

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

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

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

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).

#### Example 5

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

Make a table:

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

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

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

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

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

Here is a graph showing the two solutions:

### Review

The graphs of \begin{align*}r = 1\end{align*} and \begin{align*}r = 2 cos \theta\end{align*} are shown below.

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 \begin{align*}r = cos \theta\end{align*} and \begin{align*}r = 1 - cos \theta\end{align*}

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. \begin{align*}r = 2\end{align*} \begin{align*}r = 2 cos \theta\end{align*}
2. \begin{align*}r = sin 2 \theta\end{align*} \begin{align*} r =2 sin \theta\end{align*}
3. \begin{align*}r = 2 + 2 sin \theta\end{align*} \begin{align*} r = 2 - 2 cos \theta\end{align*}
4. \begin{align*}r = 3 cos \theta\end{align*} \begin{align*} r = 2 - cos \theta\end{align*}

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

1. \begin{align*}r_1 = csc \theta\end{align*}\begin{align*}r_2 = 2 sin \theta\end{align*}
2. \begin{align*}r_1 = cos \theta\end{align*}\begin{align*}r_2 = 1 + sin \theta\end{align*}
3. \begin{align*}r_1 = sin \theta\end{align*}\begin{align*}r_2 = sin 2 \theta\end{align*}
4. \begin{align*}r_1 = -4 sin \theta\end{align*}\begin{align*}r_2 = -4 cos \theta\end{align*}
5. \begin{align*}r_1 = 1 - 2 sin \theta\end{align*}\begin{align*}r_2 =\sqrt {9\ \mbox{cos}(\theta)}\end{align*}
6. \begin{align*}r_1 = 1 - cos \theta\end{align*}\begin{align*}r_2 = 4 cos(3\theta)\end{align*}

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### Vocabulary Language: English

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.