Math Core

Rectangular coordinates tell you how to reach a point by walking along a grid: so far right, then so far up. But a radar screen or a lighthouse beam works differently. It reports how far away something is and in which direction. That is the idea behind polar coordinates, and since a distance and an angle are exactly what a vector's magnitude and direction are, you already have the tools to use them.

Locating a point with a distance and an angle

Start at a fixed point called the pole (the origin) and draw a ray to the right called the polar axis (the positive $x$ -axis). Any point $P$ can then be described by two numbers.

Definition

Polar coordinates

A point has polar coordinates $(r, \theta)$ when it lies at a directed distance $r$ from the pole along the ray at angle $\theta$ , where $\theta$ is measured counterclockwise from the polar axis.

To plot $(3, \tfrac{2\pi}{3})$ , turn $\tfrac{2\pi}{3}$ (that's $120^\circ$ ) counterclockwise from the polar axis, then walk $3$ units out along that ray.

The point (3, 2π/3) is 3 units from the pole along the ray at 2π/3. The dashed circle shows every point with r = 3.Open in grapher →

Angles in polar coordinates are usually given in radians, but degrees work too as long as you say which you are using.

One point, many names

In rectangular coordinates every point has exactly one address. In polar coordinates every point has infinitely many.

Adding $2\pi$ to the angle brings you back to the same ray: $(3, \tfrac{2\pi}{3})$ , $(3, \tfrac{8\pi}{3})$ and $(3, -\tfrac{4\pi}{3})$ are all the same point.
A negative $r$ means walk backward: face the direction $\theta$ , then step $|r|$ units the opposite way. So $(-3, \tfrac{5\pi}{3})$ also names the point above, because the ray at $\tfrac{5\pi}{3}$ points exactly opposite the ray at $\tfrac{2\pi}{3}$ .

Equivalent polar coordinates

For any integer $n$ , the point $(r, \theta)$ is also

(r,\; \theta + 2\pi n) \qquad\text{and}\qquad (-r,\; \theta + \pi + 2\pi n).

The pole itself is $(0, \theta)$ for every $\theta$ .

Converting polar to rectangular

Picture the point $(r, \theta)$ with $r > 0$ . Dropping a perpendicular to the $x$ -axis makes a right triangle with hypotenuse $r$ and angle $\theta$ , exactly like a point on a circle of radius $r$ . So

x = r\cos\theta \qquad y = r\sin\theta.

These formulas also work when $r$ is negative, so you never need a special case.

Worked example: Polar to rectangular

Convert $(6, \tfrac{5\pi}{6})$ and $(-4, \tfrac{\pi}{4})$ to rectangular coordinates.

First point. $x = 6\cos\tfrac{5\pi}{6} = 6\left(-\tfrac{\sqrt{3}}{2}\right) = -3\sqrt{3}$ and $y = 6\sin\tfrac{5\pi}{6} = 6\left(\tfrac{1}{2}\right) = 3$ . The point is $(-3\sqrt{3}, 3)$ .

Second point. $x = -4\cos\tfrac{\pi}{4} = -4\left(\tfrac{\sqrt{2}}{2}\right) = -2\sqrt{2}$ and $y = -4\sin\tfrac{\pi}{4} = -2\sqrt{2}$ . The point is $(-2\sqrt{2}, -2\sqrt{2})$ , in Quadrant III, which makes sense for a negative $r$ with a Quadrant I angle.

Converting rectangular to polar

Going the other way, the Pythagorean theorem gives $r$ and the tangent ratio gives $\theta$ :

r = \sqrt{x^2 + y^2} \qquad \tan\theta = \frac{y}{x}.

These are the same formulas you used for a vector's magnitude and direction angle, and the same caution applies.

Common mistake

$\tan^{-1}\left(\dfrac{y}{x}\right)$ only gives angles in Quadrants I and IV. If the point has a negative $x$ -coordinate, add $\pi$ to the calculator's angle. Always check the quadrant of the original point before you trust $\theta$ .

Worked example: Rectangular to polar

Write $(-2, -2\sqrt{3})$ in polar form with $r > 0$ and $0 \le \theta < 2\pi$ .

$r = \sqrt{4 + 12} = 4$ . Then $\tan\theta = \dfrac{-2\sqrt{3}}{-2} = \sqrt{3}$ . The reference angle is $\tfrac{\pi}{3}$ , and the point is in Quadrant III, so $\theta = \pi + \tfrac{\pi}{3} = \tfrac{4\pi}{3}$ .

The polar coordinates are $(4, \tfrac{4\pi}{3})$ .

Converting equations

The same substitutions let you rewrite whole equations. Besides $x = r\cos\theta$ and $y = r\sin\theta$ , the identity $x^2 + y^2 = r^2$ is often the key.

Worked example: Polar equation to rectangular

Identify the graph of $r = 6\sin\theta$ .

Multiply both sides by $r$ so that the substitutions fit:

r^2 = 6r\sin\theta \quad\Longrightarrow\quad x^2 + y^2 = 6y.

Complete the square in $y$ : $x^2 + (y^2 - 6y + 9) = 9$ , so $x^2 + (y - 3)^2 = 9$ . The graph is a circle with center $(0, 3)$ and radius $3$ .

Rectangular equations convert just as easily. The line $x = 5$ becomes $r\cos\theta = 5$ , or $r = 5\sec\theta$ . The circle $x^2 + y^2 = 49$ becomes simply $r = 7$ . Circles centered at the pole are much simpler in polar form, which is one reason polar coordinates are worth learning.

Tip

When you multiply a polar equation by $r$ , you might add the pole as a solution. For $r = 6\sin\theta$ that's harmless, because the circle already passes through the pole at $\theta = 0$ .

Practice

Practice 1

Convert the polar point $(4, \tfrac{\pi}{6})$ to rectangular coordinates.

Enter a point like (2, -3)

Practice 2

Convert the polar point $(-2, \tfrac{\pi}{3})$ to rectangular coordinates.

Enter a point like (2, -3)

Practice 3

Which polar coordinates name the same point as $(3, \tfrac{\pi}{4})$ ?

Practice 4

Write the rectangular point $(0, -5)$ in polar form with $r > 0$ and $0 \le \theta < 2\pi$ . Give $\theta$ in radians.

Enter a point like (2, -3)

Practice 5

Write the rectangular point $(-3, 3)$ in polar form with $r > 0$ and $0 \le \theta < 2\pi$ . Give $\theta$ in radians.

Enter a point like (2, -3)

Practice 6

The polar equation $r = 6\cos\theta$ is a circle. Find its center in rectangular coordinates.

Enter a point like (2, -3)

Practice 7

Which polar equation has the same graph as the line $y = 2$ ?

Practice 8

Write $(-5, -12)$ in polar form with $r > 0$ and $0 \le \theta < 2\pi$ . Give $\theta$ in radians, rounded to the nearest hundredth.

Enter a point like (2, -3)