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Chapter 10: Problem 57

Sketch each circle in the coordinate plane and label it with both its Cartesian and polar equations. $$(x-6)^{2}+y^{2}=36$$

Short Answer

Expert verified
The circle's center is at (6, 0) with radius 6, and its polar equation is \(r = 12 \cos(\theta)\).

Step by step solution

01

Identify the elements of the Cartesian equation

The given Cartesian equation is \((x-6)^{2}+y^{2}=36\). This is the equation of a circle in standard form \((x-h)^2 + (y-k)^2 = r^2\), where \((h,k)\) is the center of the circle and \(r\) is the radius. In this case, \(h = 6\), \(k = 0\), and \(r^2 = 36\), giving a radius \(r=6\). Therefore, the circle has its center at \((6,0)\) and a radius of 6.
02

Draw the circle on the coordinate plane

On the coordinate plane, plot the center of the circle at the point \((6,0)\). Then, draw a circle around this center with a radius of 6 units.
03

Convert the Cartesian equation to polar form

To find the polar equation, we utilize the transformations: \(x = r \cos(\theta)\) and \(y = r \sin(\theta)\). Substituting \(x = r \cos(\theta)\) into the equation \((x-6)^{2}+y^{2}=36\), we have:\[(r \cos(\theta) - 6)^2 + (r \sin(\theta))^2 = 36\]. Simplifying gives:\[r^2 \cos^2(\theta) - 12r \cos(\theta) + 36 + r^2 \sin^2(\theta) = 36\]. Combining terms, \[r^2 - 12r \cos(\theta) = 0\], which can be factored as \[r(r - 12 \cos(\theta)) = 0\]. Thus, \(r = 12 \cos(\theta)\) is the polar equation.
04

Label the circle with its equations

Write the Cartesian equation \((x-6)^{2}+y^{2}=36\) and the polar equation \(r = 12 \cos(\theta)\) near the circle you have drawn on the coordinate plane to label it properly.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Cartesian Equation
A Cartesian equation is a mathematical expression that describes the location of a figure, like a circle, on the coordinate plane using the x and y axes. In this context, the given equation \((x-6)^{2}+y^{2}=36\) represents a circle.
The equation is in the standard form \((x-h)^2 + (y-k)^2 = r^2\), which makes it easy to identify the circle's key components.
  • \((h, k)\) is the center of the circle. Here, \(h = 6\) and \(k = 0\), meaning the center is at the point \( (6, 0) \).
  • \(r^2\) is the square of the radius. \(r^2 = 36\) indicates a radius \(r = 6\).
By understanding the components, you can effectively draw and work with the circle on the coordinate plane. The equation provides all the essential information about the circle's size and position.
Polar Equation
The polar equation offers another perspective for understanding a circle's position but uses polar coordinates \(r, \theta\) instead of Cartesian coordinates \(x, y\).
Polar coordinates are based on a point鈥檚 distance from a fixed center, known as the pole, and the angle \(\theta\) formed with a fixed direction.

To convert the Cartesian equation of a circle \((x-6)^{2}+y^{2}=36\) to a polar equation, we use the known transformations:
  • \(x = r \cos(\theta)\)
  • \(y = r \sin(\theta)\)
Rearranging these transformations in our circle鈥檚 equation and simplifying,
Coordinate Plane
The coordinate plane is a 2-dimensional area defined by two perpendicular lines, the x-axis (horizontal) and the y-axis (vertical). It serves as the background grid that helps us accurately graph geometric figures like circles.

To draw the circle described by the Cartesian equation \((x-6)^{2}+y^{2}=36\) on a coordinate plane:
  • Locate the center of the circle at point \( (6, 0) \).
  • From the center, measure and draw a circle with a radius of 6 units.
This graph helps to visualize the circle's position on the coordinate plane, making it intuitively clearer to see how the circle relates to other objects in the same dimensional space. The coordinate plane is an essential tool for plotting equations visually and understanding spatial relationships.

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Most popular questions from this chapter

Replace the polar equations with equivalent Cartesian equations. Then describe or identify the graph. $$r=4 \csc \theta$$

Give the foci or vertices and the eccentricities of ellipses centered at the origin of the \(x y\) -plane. In each case, find the ellipse's standard-form equation in Cartesian coordinates. Vertices: (0,卤70) Eccentricity: 0.1

Find the eccentricity of the hyperbola. Then find and graph the hyperbola's foci and directrices. $$y^{2}-x^{2}=4$$

Give the eccentricities of conic sections with one focus at the origin along with the directrix corresponding to that focus. Find a polar equation for each conic section. $$e=1, \quad y=2$$

Replace the polar equations with equivalent Cartesian equations. Then describe or identify the graph. $$\cos ^{2} \theta=\sin ^{2} \theta$$
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