Polar Coordinates — Conversion, Graphing & Equations

Polar coordinates describe a point by distance and angle instead of horizontal and vertical position. A point $(r,\theta)$ means "move $r$ units from the origin at angle $\theta$ from the positive $x$-axis."

When to Switch to Polar

Reach for polar coordinates whenever a problem leans naturally on distance from the origin or rotation around it: circles centered at the origin, spirals, orbital motion, and fields or waves that depend on distance from a central point. In Cartesian coordinates, $(3,4)$ means move $3$ along $x$ and $4$ along $y$; in polar, $(5,\theta)$ means move $5$ units out and rotate by $\theta$. That viewpoint also makes equations like $r = 2$, $r = 1 + \cos\theta$, and $r = \theta$ feel natural, because each describes distance from the origin as the angle changes. Many integrals in calculus and physics also simplify when distance and angle are the natural variables.

If a problem is built on straight horizontal/vertical structure, stay Cartesian.

The Conversion Steps

These are the relationships you apply in every conversion:

Polar to Cartesian: plug $r$ and $\theta$ into $x = r\cos\theta$ and $y = r\sin\theta$.

Cartesian to polar: first find the distance,

then choose an angle $\theta$ that points to the correct quadrant. Use $\tan\theta = \frac{y}{x}$ only when $x \ne 0$, and read the quadrant carefully, since the same tangent value appears in more than one quadrant. For example, $(-3,3)$ has $\tan\theta = -1$, but the correct angle is in Quadrant II, so $\theta = \frac{3\pi}{4}$, not $-\frac{\pi}{4}$. One special case: at the origin, $r = 0$ and any angle lands on the same point.

Full Worked Example: Convert $r = 2\cos\theta$ to Cartesian Form

This shows how a polar equation can hide a familiar graph. Start with

Multiply both sides by $r$:

Now use $r^2 = x^2 + y^2$ and $r\cos\theta = x$:

Complete the square:

So the graph is a circle centered at $(1,0)$ with radius $1$. The shape makes sense: near $\theta = 0$, $\cos\theta$ is positive and largest, so the curve extends right. When $\cos\theta$ is negative, $r$ becomes negative, flipping the point by $\pi$ and still tracing the same circle.

Where Each Step Goes Wrong (and How to Self-Check)

• Assuming a point has one polar form. It does not — $(r,\theta)$ and $(r,\theta + 2\pi)$ are the same point, as are $(r,\theta)$ and $(-r,\theta + \pi)$. Two different-looking answers can both be correct.
• Skipping the quadrant check. Using $\theta = \tan^{-1}(y/x)$ blindly can give the wrong direction even when $r$ is right. Always confirm the quadrant matches the signs of $x$ and $y$.
• Mixing radians and degrees. The graph depends on the unit your problem uses, so keep it consistent throughout.
• Misreading negative $r$. It does not mean "invalid." It means move in the opposite direction of the given angle. At the origin, the reverse trap appears: do not force a single angle, because none is required there.

Try the Pattern Yourself

Convert $r = 4\sin\theta$ to Cartesian form and identify the graph. If you land on a circle, you are seeing the same conversion pattern pointed in a slightly different direction — multiply by $r$, substitute, then complete the square.