Problem 69 - Calculus Early Transcendentals

Question

Show that the polar equation \(r=a \sin \theta+b \cos \theta,\) where \(a b \neq 0,\) represents a circle, and find its center and radius.

Step-by-Step Solution

Answer

The polar equation represents a circle centered at \((b, a)\) with radius \(r = \sqrt{a^2 + b^2}\).

Step 1: Recognize the Polar Equation

The given polar equation is \(r = a \sin \theta + b \cos \theta\). This form suggests that the equation might be representing a conic section, potentially a circle.

Step 2: Convert to Cartesian Coordinates

To convert the polar equation into Cartesian coordinates, use the relationships: \(x = r \cos \theta\), \(y = r \sin \theta\), and \(r = \sqrt{x^2 + y^2}\). Substitute these into the polar equation: \(\sqrt{x^2 + y^2} = a \frac{y}{r} + b \frac{x}{r}\). Simplify to get \(r = a y + b x\).

Step 3: Square Both Sides

Square both sides of the equation \(r = a y + b x\) to eliminate the square root: \(x^2 + y^2 = (a y + b x)^2\). This results in \(x^2 + y^2 = a^2 y^2 + 2abxy + b^2 x^2\).

Step 4: Simplify and Rearrange the Equation

Rearrange the equation \(x^2 + y^2 = a^2 y^2 + 2abxy + b^2 x^2\) into a more familiar cartesian form: \((x^2 + y^2) - b^2 x^2 - a^2 y^2 = 2abxy\). Factor the left side: \((x^2 - b^2 x^2) + (y^2 - a^2 y^2) = 0\), which simplifies further as \((1-b^2)x^2 + (1-a^2)y^2 = 0\). This form will help in identifying if the equation is a circle.

Step 5: Complete the Square

In order to complete the square, set \((x^2 - xx) + (y^2 - yy) = rr\) as needed. Rewrite the equation \((x - b)^2 + (y - a)^2 = b^2 + a^2\) by balancing the coefficients. This step will allow us to find the center and radius of the circle.

Step 6: Identify Center and Radius

Recognize the equation \((x - b)^2 + (y - a)^2 = r^2\) as a circle with its center at \((b, a)\) and radius \(r = \sqrt{a^2 + b^2}\). Thus, the given polar equation represents a circle with these characteristics.

Key Concepts

Polar Equations

Polar equations express the relationships between points in the polar coordinate system. Instead of using the Cartesian coordinates \((x, y)\), we use \((r, \theta)\), where \(r\) is the distance from the origin and \(\theta\) is the angle from the positive x-axis. Polar equations are particularly useful in describing curves like circles and spirals.
In the problem, the given polar equation is \(r = a \sin \theta + b \cos \theta\). This form hints at a conic section, and in particular, it can represent a circle under appropriate conditions. The equation combines trigonometric functions of \(\sin\) and \(\cos\), which ties directly into how points are defined in polar coordinates.

Circle Properties

A circle is defined as the set of all points that are equidistant from a central point. In our derived Cartesian equation \((x - b)^2 + (y - a)^2 = r^2\), it is easy to observe the circle properties.

• The center of the circle is \((b, a)\), indicating the horizontal and vertical positions.
• The radius \(r\) is the fixed distance from any point on the circle to its center.

Understanding this helps in realizing why any manipulation like completing the square (as shown in the solution) can transform a general quadratic into the standard circle equation, revealing its center and radius.

Cartesian Coordinates Conversion

The process of converting from polar to Cartesian coordinates involves a set of straightforward trigonometric and algebraic relations.

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

Substituting these into our equation aligns the polar definition with the Cartesian plane.
Through manipulation and substitution, it's possible to convert polar equations to a Cartesian form that is more recognizable and easier to manipulate algebraically. This is significant because many geometric shapes, like circles, have simple representations in Cartesian form, allowing for easy calculation of properties like centers and radii.