91Ó°ÊÓ

Polar coordinates represent points on a plane using a radius and an angle. Here, the radius, denoted as \( r \), is the distance of the point from the origin, while \( \theta \) is the angle the line makes with the positive x-axis.
This system is particularly useful in scenarios involving circular or spiral shapes, where angles and distances provide a more natural description than straight horizontal and vertical distances. In the exercise example, we start with the polar equation \( r = 6 \cos \theta \). This tells us that the radius varies with the angle \( \theta \), creating a specific curve when plotted.
Understanding polar coordinates involves recognizing that they give a different but equally valid way to locate points in a plane, contrasting with the more common rectangular coordinates.

Rectangular coordinates, also known as Cartesian coordinates, describe points using horizontal and vertical distances from a reference point, typically called the origin. Here, the position is given by \( (x, y) \), where \( x \) is the horizontal distance and \( y \) is the vertical distance from the origin.
Converting equations from polar to rectangular coordinates involves using known relationships:

• \( x = r \cos \theta \)
• \( y = r \sin \theta \)
• \( r^2 = x^2 + y^2 \)
• \( \cos \theta = \frac{x}{r} \)

By substituting these relationships into a polar equation, you can translate it to the rectangular form.
In our task, using these conversion formulas, we transform \( r = 6 \cos \theta \) into \( x^2 + y^2 = 6x \), setting the stage for further manipulation.

Completing the square is a technique to transform a quadratic equation into a more easily interpretable form, such as identifying a geometric object. This involves making the quadratic term a perfect square trinomial.
In our transformed equation \( x^2 - 6x + y^2 = 0 \), we will focus on the \( x \) terms. We aim to rewrite it as \( (x-3)^2 + y^2 = 9 \). Here’s how it’s done:

• Take the coefficient of \( x \), which is \(-6\), halve it to get \(-3\), then square this result to get \(9\).
• Add and subtract \( 9 \) within the equation, allowing us to write \( x^2 - 6x + 9 \) as \( (x-3)^2 \).

Completing the square aligns the equation with the standard form of a circle, making the geometry evident.

Graphing circles can become straightforward once the equation of a circle is in its standard form. This form is usually \( (x-h)^2 + (y-k)^2 = r^2 \) where \( (h, k) \) represents the circle's center and \( r \) is the radius.
For the exercise, after completing the square, the equation \( (x-3)^2 + y^2 = 9 \) tells us:

• The circle has a center at \( (3, 0) \).
• The radius is \( 3 \), since \( 9 \) is the square of radius \( (3^2) \).

This information lets you accurately graph the circle by marking the center and counting out from there. With the center known, draw the circle using the radius, capturing its full size and position on the coordinate plane. This geometric insight is another benefit of converting polar to rectangular coordinates.