Chapter 10: Problem 31
Find a polar equation that has the same graph as the equation in \(x\) and \(y\). $$(x-1)^{2}+y^{2}=1$$
Short Answer
Expert verified
The polar equation is \(r = 2\cos\theta\).
Step by step solution
01
Identify Circle Properties
The given equation \((x-1)^{2} + y^{2} = 1\) represents a circle with center \(1, 0\) and radius \(1\).
02
Convert to Polar Coordinates
In polar coordinates, \(x = r\cos\theta\) and \(y = r\sin\theta\). Substitute these into the original equation: \((r\cos\theta - 1)^{2} + (r\sin\theta)^{2} = 1\).
03
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 \((\cos^2\theta + \sin^2\theta = 1)\).
04
Combine and Rearrange
Combine terms: \(r^2 - 2r\cos\theta + 1 = 1\). Simplify this to obtain: \(r^2 - 2r\cos\theta = 0\).
05
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 \(r = 2\cos\theta\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
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 \(1\).
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:
- \(x = r\cos\theta\)
- \(y = r\sin\theta\)
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 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 \(\sqrt{1} = 1\).