Problem 51 - Algebra and Trigonometry

Question

Show that the graph of \(r=a \cos \theta+b \sin \theta\) is a circle, and find its center and radius.

Step-by-Step Solution

Answer

The circle's center is \(\left(\frac{a}{2}, \frac{b}{2}\right)\) and radius is \(\frac{\sqrt{a^2 + b^2}}{2}\).

Step 1: Rewrite the Polar Equation

The exercise asks us to demonstrate that the graph of the polar equation \( r = a \cos \theta + b \sin \theta \) is a circle. To do this, let's start by changing the polar equation into a Cartesian form. From polar coordinates, we know that \( x = r \cos \theta \) and \( y = r \sin \theta \). Rearrange for \( r \) in terms of \( x \) and \( y \), we have \( r^2 = x^2 + y^2 \). The polar equation is given as \( r = a \cos \theta + b \sin \theta \), multiplying through by \( r \), we obtain \( r^2 = ar \cos \theta + br \sin \theta \).

Step 2: Substitution for x and y

Substitute \( x = r \cos \theta \) and \( y = r \sin \theta \) into the equation derived in Step 1: \( x = ar \cos^2 \theta + br \cos \theta \sin \theta \) and \( y = ar \cos \theta \sin \theta + br \sin^2 \theta \). Substitute these into \( r^2 = ar \cos \theta + br \sin \theta \), resulting in the equation \( x^2 + y^2 = ax + by \).

Step 3: Completing the Square

To show that this is an equation of a circle, let's complete the square. Rearrange the terms \( x^2 - ax + y^2 - by = 0 \) into \((x^2 - ax) + (y^2 - by) = 0\). Now complete the square for each group: \((x - \frac{a}{2})^2 - (\frac{a^2}{4})\) and \((y - \frac{b}{2})^2 - (\frac{b^2}{4})\). Simplifying gives \((x - \frac{a}{2})^2 + (y - \frac{b}{2})^2 = \frac{a^2}{4} + \frac{b^2}{4}\).

Step 4: Identify the Center and Radius

The equation \((x - \frac{a}{2})^2 + (y - \frac{b}{2})^2 = \frac{a^2 + b^2}{4}\) is the standard form of a circle, \((x - h)^2 + (y - k)^2 = r^2\), where \((h, k)\) is the center and \(r\) is the radius. Thus, the center of the circle is \(\left(\frac{a}{2}, \frac{b}{2}\right)\) and the radius is \(\sqrt{\frac{a^2 + b^2}{4}} = \frac{\sqrt{a^2 + b^2}}{2}\).

Key Concepts

Cartesian Form

The transition from polar coordinates to Cartesian form is an important mathematical process. When working with polar equations, such as \( r = a \cos \theta + b \sin \theta\), converting into the Cartesian form allows us to better visualize and understand the equation on a conventional x-y plane.
Polar coordinates rely on an angle \(\theta\) and radius \(r\) to define a position, whereas Cartesian coordinates use x and y values.To convert, we use the relationships:

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

Replacing \(r\cos\theta\) and \(r\sin\theta\) in the polar equation with x and y, we utilize these conversions.
For instance, in the problem, the step-by-step solution involves substituting \(\cos\theta\) and \(\sin\theta\) to transform the equation into \(x^2 + y^2 = ax + by\), a Cartesian equation. This sets the stage for further simplification, allowing us to express it in more familiar terms.

Completing the Square

Completing the square is a method used to manipulate quadratic expressions into perfect square trinomials.
It is particularly useful when identifying features of quadratic forms, like the center and radius of a circle.In our context, after transforming the equation into Cartesian coordinates, we reach:

• \((x^2 - ax) + (y^2 - by) = 0\)

To complete the square, the terms \(x^2 - ax\) and \(y^2 - by\) need rearranging into perfect squares:

• \((x - \frac{a}{2})^2 - (\frac{a^2}{4})\)
• \((y - \frac{b}{2})^2 - (\frac{b^2}{4})\)

Here’s the simplification: you add and subtract \(\frac{a^2}{4}\) and \(\frac{b^2}{4}\) inside each respective bracket to maintain balance.
This operation eventually leads us to the circle's standard form, revealing key properties such as center and radius.
Completing the square is an elegant tool that untangles complex expressions into recognizable shapes.

Equation of a Circle

A circle's equation is pivotal in understanding its geometric properties. The standard form in the Cartesian plane is:\[(x - h)^2 + (y - k)^2 = r^2\]Where \((h, k)\) represents the circle's center and \(r\) stands for its radius.
In our example, after converting from polar to Cartesian form and completing the square, we find:\[(x - \frac{a}{2})^2 + (y - \frac{b}{2})^2 = \left( \frac{\sqrt{a^2 + b^2}}{2} \right)^2\]The coordinates \(\left(\frac{a}{2}, \frac{b}{2}\right)\) pinpoint the center, and the radius is \(\frac{\sqrt{a^2 + b^2}}{2}\).
This equation allows us to immediately see the circle's center and radius way beyond the algebraic complexity.

• The center \((h, k)\) is derived from the terms \(x - \frac{a}{2}\) and \(y - \frac{b}{2}\).
• The expression \(\frac{a^2 + b^2}{4}\) under the square root determines the radius.

Recognizing such forms in different coordinates and manipulating them is fundamental in linking algebra with geometry.