91Ó°ÊÓ

Polar coordinates offer a unique way to describe the position of a point in a plane. Instead of using horizontal and vertical distances, as in Cartesian coordinates, polar coordinates use a radius and an angle.
In polar coordinates:

• The variable \( r \) signifies the distance from the point to the origin (or pole).
• The angle \( \theta \) describes the direction of the point from the positive x-axis.

When you write a point in polar coordinates, it usually takes the form \((r, \theta)\). This system is particularly advantageous for dealing with curves that have rotational symmetry, such as circles and spirals. It simplifies equations and makes certain problems easier to visualize.
For the exercise given, the polar equation \( r - 5 \cos \theta = 0 \) easily describes positions relative to the origin, relying on the cosine function to determine the angle \( \theta \).

Cartesian coordinates are the most familiar form of describing a point in a plane. They use an x-coordinate and a y-coordinate to specify a location relative to two perpendicular number lines – the x-axis and the y-axis.
The form \((x, y)\) represents any point in the Cartesian plane.

• x is the horizontal distance from the vertical y-axis.
• y is the vertical distance from the horizontal x-axis.

For conversion from polar to Cartesian coordinates, utilize the relationships:

• \( x = r \cos \theta \)
• \( y = r \sin \theta \)

In this exercise, by converting the polar equation \( r = 5 \cos \theta \) first into \( r^2 = x^2 + y^2 \), then simplifying further by substituting \( \cos \theta = \frac{x}{r} \), we derive the Cartesian form \( x^2 + y^2 = 5x \). This transformation enables us to interpret the position on a plane using more familiar x and y terms.

The equation of a circle in its standard Cartesian form is a familiar concept. It is given by the formula: \[(x - h)^2 + (y - k)^2 = r^2\]where:

• \( (h, k) \) are the coordinates of the circle's center.
• \( r \) is the radius of the circle.

This equation stems from the definition that all points are equidistant from the center. Points satisfying this equation lie on the circle.
In the context of the exercise, our final Cartesian equation \( (x - \frac{5}{2})^2 + y^2 = \frac{25}{4} \) represents a circle. Here, the center is at \( (\frac{5}{2}, 0) \) and the radius is \( \frac{5}{2} \). Completing the square helped in reformulating \( x^2 + y^2 = 5x \) into this standard form, making the characteristics of the circle more apparent.