/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 44 Graph each circle using a graphi... [FREE SOLUTION] | 91影视

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Chapter 8: Problem 44

Graph each circle using a graphing calculator. Use a square viewing window. Give the domain and range. $$(x+2)^{2}+(y+3)^{2}=36$$

Short Answer

Expert verified
The domain is \([-8, 4]\) and the range is \([-9, 3]\).

Step by step solution

01

Identify the Center and Radius of the Circle

The equation of a circle in standard form is \((x-h)^2 + (y-k)^2 = r^2\), where \((h, k)\) is the center and \(r\) is the radius. For the given equation \((x+2)^2 + (y+3)^2 = 36\), we can identify: \(h = -2\), \(k = -3\), and \(r^2 = 36\). Thus, the center is \((-2, -3)\) and the radius \(r = \sqrt{36} = 6\).
02

Graph the Circle

Using a graphing calculator, input the equation \((x+2)^2 + (y+3)^2 = 36\). Select a 'square' viewing window to ensure equal scaling on both axes. A suitable window might be from \(-10\) to \(10\) for both \(x\) and \(y\) axes. This allows the entire circle to be visible with the center at \((-2, -3)\).
03

Determine the Domain of the Circle

The domain of a circle, given its equation and center, is the interval of \(x\)-values it covers. With center at \((-2, -3)\) and radius \(6\), the farthest points on the \(x\)-axis are \(x = -2 - 6 = -8\) and \(x = -2 + 6 = 4\). Thus, the domain is \([-8, 4]\).
04

Determine the Range of the Circle

Similarly, the range is the interval of \(y\)-values covered by the circle. With center at \((-2, -3)\) and radius \(6\), the furthest points on the \(y\)-axis are \(y = -3 - 6 = -9\) and \(y = -3 + 6 = 3\). Thus, the range is \([-9, 3]\).

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Key Concepts

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

Equation of a Circle
In mathematics, the equation of a circle provides a formula that defines all the points on the perimeter of the circle. The standard form of a circle's equation is given by \[(x - h)^2 + (y - k)^2 = r^2\]where:
  • \((h, k)\) represents the coordinates of the circle's center.
  • \(r\) represents the radius.
Each point \((x, y)\) that satisfies the equation is \(r\) units away from the center. For the circle in the exercise, \((x+2)^2 + (y+3)^2 = 36\), the center is \((-2, -3)\) and the radius is \(6\), since \(r^2 = 36\).This form makes it easy to identify both the center and the radius, crucial for graphing and understanding the circle's properties.
Domain and Range
The domain and range of a circle are essential concepts when graphing and analyzing its position on the Cartesian plane.
  • The **domain** refers to the set of possible values for the variable \(x\) that a circle covers. It can be derived from finding how far the circle extends horizontally. For a circle with a center at \((-2, -3)\) and a radius of \(6\), its furthest horizontal reach is \(-8\) (left) to \(4\) (right), thus giving the domain \([-8, 4]\).
  • The **range** indicates the possible values for \(y\) the circle covers vertically. Using the same center and radius, the topmost and bottommost points are \(3\) and \(-9\), respectively, resulting in the range \([-9, 3]\).
Understanding these intervals ensures that the graph effectively represents all possible circle coordinates.
Graphing Calculator
A graphing calculator is an excellent tool for visualizing mathematical relationships, like those found in circles. You can use it to graph the circle's equation, \((x+2)^2 + (y+3)^2 = 36\), and see how it is positioned on the graph. When inputting this equation into a graphing calculator, remember these simple steps:
  • Ensure the calculator is set to graph mode.
  • Enter the equation in its standard form.
  • Adjust the viewing window to equal dimensions horizontally and vertically. A square window, such as \([-10, 10]\) on both axes, will correctly display the circle without distortion.
A graphing calculator provides a visual means to confirm calculations and understand the circle's geometry. It helps check the center and radius by visually verifying their effect on the circle鈥檚 position and size in the window.
Circle Center and Radius
Locating the center and radius of a circle is fundamental in understanding its size and position. When dealing with the standard circle equation, \((x-h)^2 + (y-k)^2 = r^2\), these constants naturally emerge:
  • The **center** (\(h, k\)) indicates the point right in the middle of the circle, crucial for any transformations or translations of the circle on the graph.
  • The **radius** (\(r\)) gives the fixed distance from the center to any point on the circle. It greatly influences how large or small the circle will appear on the graph.
For instance, in the given equation \((x+2)^2 + (y+3)^2 = 36\), we extract \(h = -2, k = -3,\) and \(r = 6\). Each of these values plays a role in the circle's visual representation and measurements, allowing you to accurately depict and describe the circle's shape and position.

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Most popular questions from this chapter

Solve each nonlinear system of equations analytically for all real solutions. $$\begin{aligned} x^{2}+y^{2} &=4 \\ 5 x^{2}+5 y^{2} &=28 \end{aligned}$$

Use substitution to solve the nonlinear system of equations in three variables. Note that solutions are ondered triples. $$\begin{aligned} &\frac{x^{2}}{16}+\frac{y^{2}}{4} \leq 1\\\ &x^{2}-y^{2} \geq 1 \end{aligned}$$

Write the equation in standard form for a hyperbola centered at ( \(h, k\) ). Identify the center and vertices. $$4 x^{2}+16 x-9 y^{2}+18 y=29$$

Use substitution to solve the nonlinear system of equations in three variables. Note that solutions are ondered triples. $$\begin{aligned} 2 x^{2}+y^{2}+3 z^{2} &=3 \\ 2 x+y-z &=1 \\ x+y &=0 \end{aligned}$$

Equation of an Ellipse Use the definition of an ellipse to find an equation of an ellipse with foci \((3,0)\) and \((-3,0),\) where the sum of the distances from any point of the ellipse to the two foci is \(10 .\)
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