You are working in Art Class one afternoon and decide to draw the "Olympic Rings". These are a set of rings that lock together and are the symbol of the Olympics.

You can do this on the computer by generating a circle for each of the rings using an equation in polar coordinates. The equations you use for the first two rings are \begin{align*}r = 2\sin \theta\end{align*}

### Intersections of Polar Curves

When you worked with a system of linear equations with two unknowns, finding the point of intersection of the equations meant finding the coordinates of the point that satisfied both equations. If the equations are rectangular equations for curves, determining the point(s) of intersection of the curves involves solving the equations algebraically since each point will have one ordered pair of coordinates associated with it.

Let's take a look at a few problems that involve intersections of polar curves.

1. Solve the following system of equations algebraically:

\begin{align*}x^2 + 4y^2 - 36 & = 0 \\
x^2 + y & = 3\end{align*}

Before solving the system, graph the equations to determine the number of points of intersection.

The graph of \begin{align*}x^2 + 4y^2 - 36 = 0\end{align*}

\begin{align*}& x^2 + 4y^2 - 36 = 0 \to x^2 + 4y^2 = 36 && x^2 + 4y^2 + 0y = 36 && \quad x^2 + 4y^2 + 0y = 36 \\
& x^2 + y = 3 \to x^2 + 0y^2 + y = 3 && -1(x^2 + 0y^2 + y = 3) && -x^2 - 0y^2 - y = -3 \\
&&& && \quad \underline{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;} \\
&&& && \quad 4y^2 - y = 33\end{align*}

Using the quadratic formula, \begin{align*}a = 4 \ b = -1 \ c = -33\end{align*}

\begin{align*}y & = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(4)(-33)}}{2(4)} \\
y & = \frac{1 + 23}{8} = 3 \qquad y = \frac{1 - 23}{8} = -2.75\end{align*}

These values must be substituted into one of the original equations.

\begin{align*}x^2 + y &= 3 && \qquad \ \ x^2 + y = 3 \\
x^2 +3 & = 3 && x^2 + (-2.75) = 3 \\
x^2 & = 0 && \qquad \qquad \ \ x^2 = 5.75 \\
x & = 0 && x = \pm \sqrt{5.75} \ \approx 2.4 \end{align*}

The three points of intersection as determined algebraically in Cartesian representation are \begin{align*}\text{A} (0, 3), \text{B} (2.4, -2.75)\end{align*}

If we are working with polar equations to determine the polar coordinates of a point of intersection, we must remember that there are many polar coordinates that represent the same point. Remember that switching to polar form changes a great deal more than the notation. Unlike the Cartesian system which has one name for each point, the polar system has an infinite number of names for each point. One option would be to convert the polar coordinates to rectangular form and then to convert the coordinates for the intersection points back to polar form. Perhaps the best option would be to explore some more examples. As these problems are presented, be sure to use your graphing calculator to create your own visual representations of the equations presented.

2. Determine the polar coordinates for the intersection point(s) of the following polar equations: \begin{align*}r = 1\end{align*}

Begin with the graph. Using the process described in the technology section in this chapter; create the graph of these polar equations on your graphing calculator. Once the graphs are on the screen, use the **trace** function and the arrow keys to move the cursor around each graph. As the cursor is moved, you will notice that the equation of the curve is shown in the upper left corner and the values of \begin{align*}\theta, x, y\end{align*}

\begin{align*}& r = 1 && 2 \cos \theta = 1 \\
& r = 2 \cos \theta && \cos \theta = \frac{1}{2} \\
& && \cos^{-1}(\cos \theta) = \cos^{-1} \frac{1}{2} \end{align*}

\begin{align*}\theta = \frac{\pi}{3}\end{align*}

The points of intersection are \begin{align*}\left(1, \frac{\pi}{3} \right)\end{align*} and \begin{align*}\left(1, \frac{5 \pi}{3} \right)\end{align*}. However, these two solutions only cover the possible values \begin{align*}0 \le \theta \le 2 \pi\end{align*}. If you consider that \begin{align*}\cos \theta = \frac{1}{2}\end{align*} is true for an infinite number of theta these solutions must be extended to include \begin{align*}\left (1 , \frac{\pi}{3} \right )\end{align*} and \begin{align*}\left (1, \frac{5 \pi}{3} \right ) + 2 \pi k, k \varepsilon Z.\end{align*} Now the solutions include all possible rotations.

3. Find the intersection of the graphs of \begin{align*}r = \sin \theta\end{align*} and \begin{align*}r = 1 - \sin \theta\end{align*}

Begin with the graph. You can create these graphs using your graphing calculator.

\begin{align*}& r = \sin \theta && \quad \ \sin \theta = 1 - \sin \theta \\ & r = 1 - \sin \theta && \ \ 2 \sin \theta = 1 \\ & && \quad \sin \theta \ = \frac{1}{2}\\ & r = \sin \theta && \theta = \frac{\pi}{6} \ \text{in the first quadrant and}\ \theta = \frac{5 \pi}{6} \ \text{in the second quadrant.}\\ & r = \sin \frac{\pi}{6} && \text{The intersection points are}\ \left ( \frac{1}{2}, \frac{\pi}{6} \right ) \ \text{and} \ \left ( \frac{1}{2}, \frac{5 \pi}{6} \right )\\ & r = \frac{1}{2} && \text{Another intersection point seems to be the origin} \ (0, 0).\end{align*}

If you consider that \begin{align*}\sin \theta = \frac{1}{2}\end{align*} is true for an infinite number of theta as was \begin{align*}\cos \theta = \frac{1}{2}\end{align*} in #2, the same consideration must be applied to include all possible solutions. To prove if the origin is indeed an intersection point, we must determine whether or not both curves pass through (0, 0).

\begin{align*}& r = \sin \theta && r = 1 - \sin \theta \\ & 0 = \sin \theta && 0 = 1 - \sin \theta \\ & r = 0 && 1 = \sin \theta \\ &&& \frac{\pi}{2} = 0\end{align*}

From this investigation, the point (0, 0) was on the curve \begin{align*}r = \sin \theta\end{align*} and the point \begin{align*} \left ( 0, \frac{ \pi}{2} \right )\end{align*} was on the curve \begin{align*}r = 1 - \sin \theta\end{align*}. Because the second coordinates are different, it seems that they are two different points. However, the coordinates represent the same point (0,0). The intersection points are \begin{align*} \left( \frac{1}{2}, \frac{\pi}{6} \right), \left( \frac{1}{2}, \frac{5 \pi}{6} \right )\end{align*} and (0,0).

Sometimes it is helpful to convert the equations to rectangular form, solve the system and then convert the polar coordinates back to polar form.

### Examples

#### Example 1

Earlier, you were asked to find the intersection of two equations.

The goal is the find the place where the two equations meet. Therefore, you want the point where they are equal. Mathematically, this is:

\begin{align*} 2\cos \theta = 2\sin \theta\\ \cos \theta = \sin \theta\\ \frac{\sin \theta}{\cos \theta} = 1\\ \tan \theta = 1\\ \theta = 45^\circ, 225^\circ\\ \end{align*}

#### Example 2

Find the intersection of the graphs of \begin{align*}r = \sin 3 \theta\end{align*} and \begin{align*}r = 3 \sin \theta\end{align*}.

There appears to be one point of intersection.

\begin{align*}r & = \sin 3 \ \theta && r = 3 \sin \theta \\ 0 & = \sin 3 \ \theta && 0 = 3 \sin \theta \\ 0 & = \theta && 0 = \sin \theta \\ &&& 0 = \theta\end{align*}

**The point of intersection is** (0, 0)

#### Example 3

Find the intersection of the graphs of \begin{align*}r = 2 + 2 \sin \theta\end{align*} and \begin{align*}r = 2 - 2 \cos \theta\end{align*}

\begin{align*}r & = 2 + 2 \sin \theta && r = 2 + 2 \sin \theta \\ r & = 2 + 2 \sin \left( \frac{3 \pi}{4} \right) && r = 2 + 2 \sin \frac{7 \pi}{4} \\ r & \approx 3.4 && r \approx 0.59 \\ r & = 2 + 2 \sin \theta && r = 2 - 2 \cos \theta \\ 0 & = 2 + 2 \sin \theta && 0 = 2 - 2 \cos \theta \\ -1 & = \sin \theta && 1 = \cos \theta \\ \theta & = \frac{3\pi}{2} && \theta = 0\end{align*}

Since both equations have a solution at \begin{align*}r = 0\end{align*} , that is (0, \begin{align*}\frac{3\pi}{2}\end{align*}) and (0, 0), respectively, and these two points are equivalent, the two equations will intersect at (0, 0).

\begin{align*}r & = 2 + 2 \sin \theta \\ r & = 2 - 2 \cos \theta \\ 2 + 2 \sin \theta & = 2 - 2 \cos \theta \\ 2 \sin \theta & = -2 \cos \theta \\ \frac{2 \sin \theta}{2 \cos \theta} & = -\frac{2 \cos \theta}{2 \cos \theta} \\ \frac{\sin \theta}{\cos \theta} & = -1 \\ \tan \theta & = -1 \\ \theta & = \frac{3 \pi}{4} \ \text{and} \ \theta = \frac{7 \pi}{4}\end{align*}

The points of intersection are \begin{align*} \left(3.4, \frac{3 \pi}{4} \right), \left(0.59, \frac{7 \pi}{4} \right)\end{align*} and \begin{align*}(0, 0)\end{align*}.

#### Example 4

Determine two polar curves that will never intersect.

There are several answers here. The most obvious are any two pairs of circles, for example \begin{align*}r = 3\end{align*} and \begin{align*}r = 9\end{align*}.

### Review

Find all points of intersection for each of the following pairs of graphs. Answers should be in polar coordinates with \begin{align*}0\leq \theta < 2\pi\end{align*}.

- \begin{align*}r=2\end{align*} and \begin{align*}r=2\cos \theta\end{align*}
- \begin{align*}r=3\end{align*} and \begin{align*}r=3\sin \theta\end{align*}
- \begin{align*}r=1\end{align*} and \begin{align*}r=2\sin \theta\end{align*}
- \begin{align*}r=\sin \theta\end{align*} and \begin{align*}r=1 + \sin \theta\end{align*}
- \begin{align*}r=\sin 2\theta \end{align*} and \begin{align*}r=2\sin \theta\end{align*}
- \begin{align*}r=3+3\sin \theta\end{align*} and \begin{align*}r=3-3\cos \theta\end{align*}
- \begin{align*}r=\cos 3 \theta\end{align*} and \begin{align*}r=\sin 3\theta\end{align*}
- \begin{align*}r=\sin 2 \theta\end{align*} and \begin{align*}r=\sin 3 \theta\end{align*}
- \begin{align*}r=3 \cos \theta\end{align*} and \begin{align*}r=2-\cos \theta\end{align*}
- \begin{align*}r=\cos \theta\end{align*} and \begin{align*}r=1-\cos \theta\end{align*}
- \begin{align*}r=3\sin \theta\end{align*} and \begin{align*}r=2-\sin \theta\end{align*}
- \begin{align*}r^2=\sin \theta\end{align*} and \begin{align*}r^2=\cos \theta\end{align*}
- \begin{align*}r^2=\sin 2\theta\end{align*} and \begin{align*}r^2=\cos 2\theta\end{align*}
- Explain why one point of intersection of polar graphs can be represented by an infinite number of polar coordinate pairs.
- Explain why the point (0,0) in polar coordinates is not always found as a point of intersection when solving for points of intersection algebraically.

### Review (Answers)

To see the Review answers, open this PDF file and look for section 6.7.