Problem 31 - Precalculus: Functions and Graphs

Question

Find a polar equation that has the same graph as the equation in \(x\) and \(y\). $$(x-1)^{2}+y^{2}=1$$

Step-by-Step Solution

Answer

The polar equation is \(r = 2\cos\theta\).

Step 1: Identify Circle Properties

The given equation \((x-1)^{2} + y^{2} = 1\) represents a circle with center \(1, 0\) and radius \(1\).

Step 2: Convert to Polar Coordinates

In polar coordinates, \(x = r\cos\theta\) and \(y = r\sin\theta\). Substitute these into the original equation: \((r\cos\theta - 1)^{2} + (r\sin\theta)^{2} = 1\).

Step 3: Simplify the Polar Equation

Expand the equation: \((r^2\cos^2\theta - 2r\cos\theta + 1) + r^2\sin^2\theta = 1\). Notice that \(r^2(\cos^2\theta + \sin^2\theta) = r^2\) because \((\cos^2\theta + \sin^2\theta = 1)\).

Step 4: Combine and Rearrange

Combine terms: \(r^2 - 2r\cos\theta + 1 = 1\). Simplify this to obtain: \(r^2 - 2r\cos\theta = 0\).

Step 5: Factorize and Solve for r

Factorize the equation: \(r(r - 2\cos\theta) = 0\). Thus, we have \(r = 0\) or \(r = 2\cos\theta\). A circle in polar coordinates corresponding to the given equation is \(r = 2\cos\theta\).

Key Concepts

Circle Equations

Understanding circle equations is essential for analyzing geometric shapes. A circle in a two-dimensional space is defined by its center and its radius. The standard equation of a circle in Cartesian coordinates, where the center is at point \((a, b)\) and the radius is \(r\), can be expressed as: \[(x - a)^2 + (y - b)^2 = r^2\]In our given exercise, the circle's equation is \((x - 1)^2 + y^2 = 1\), which means:

• The center of the circle is located at \((x, y) = (1, 0)\).
• The radius of the circle is \(1\).

This equation tells us everything we need to start plotting or converting the circle into other coordinate systems, such as polar coordinates.

Cartesian to Polar Conversion

Converting equations from Cartesian to polar coordinates allows for a different view of geometric shapes, often simplifying into functions of angle \(\theta\). In polar coordinates:

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

Replace every instance of \(x\) and \(y\) in the Cartesian equation with these expressions to transform into a polar equation. In our problem, start with the substitution: \[(r\cos\theta - 1)^2 + (r\sin\theta)^2 = 1\]Next, expand and simplify the equation. We use the identity: \(\cos^2\theta + \sin^2\theta = 1\), simplifying it further into \(r^2 - 2r\cos\theta = 0\). By converting to polar, the equation captures how the radius \(r\) changes with \(\theta\), showing how circles can be described by angle-based systems.

Radius and Center of Circle

Identifying the radius and center of a circle is crucial for understanding its properties. With practice, you can quickly discern these from a standard circle equation. Let's see how to extract these values from our exercise.
The equation \((x-1)^2 + y^2 = 1\) gives us this data:

• The shift in the equation \((x-1)\) indicates the center is shifted along the x-axis to \(x = 1\).
• The lack of a shift for \(y\) (no \(b\) term subtracted) tells us \(y = 0\) for the center.
• The right side of the equation, \(1\), is the square of the circle's radius, hence the radius is \(\sqrt{1} = 1\).

Having the center and radius lets you graph circles easily or transition the equation into other forms. By mastering these identifiers, working with circles becomes straightforward, whether analytically or graphically.