Suppose that a point O in the plane is chosen as origin, and let Ox be a directed line through O, with a given unit of length. For any point P in the plane, let r =  |OP| and, if P is not O, let θ‎ be the angle (in radians) that OP makes with Ox, the angle being given a positive sense anticlockwise from Ox. The angle θ‎ satisfies 0 ≤ θ‎  <  2π‎. Then (r,θ‎) are the polar coordinates of P.

The polar coordinates of P

Suppose that Cartesian coordinates are taken with the same origin and the same unit of length, with positive x-axis along the directed line Ox. Then the Cartesian coordinates (x,y) of a point P can be found from (r,θ‎), by x =  rcosθ‎, y = rsinθ‎. Conversely, the polar coordinates can be found from the Cartesian coordinates by $r=\sqrt{x^2+y^2}$, and θ‎ is such that

(Note that θ‎ = tan⁻¹(y/x) but this is not, of itself, sufficient to uniquely determine θ‎.) In certain circumstances, authors may allow r to be negative, in which case the polar coordinates (r,θ‎) give the same point as (–r,θ‎ + π‎).

In mechanics, it is useful, when a point P has polar coordinates (r,θ‎), to define unit vectors e_r and e_θ‎ by

where i and j are unit vectors in the directions of the positive x- and y-axes. Then e_r is a unit vector along OP in the direction of increasing r, and e_θ‎ is a unit vector perpendicular to this in the direction of increasing θ‎. These vectors satisfy e_r∙e_θ‎ =0 and e_r×e_θ‎ =k, where k =  i×j. See radial and transverse components.