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Chapter 3: Problem 24

Exer. 23-34: Sketch the graph of the circle or semicircle. $$ x^{2}+y^{2}=7 $$

Short Answer

Expert verified
The circle is centered at (0, 0) with a radius of \(\sqrt{7}\).

Step by step solution

01

Understanding the Equation

The given equation is \(x^2 + y^2 = 7\). This is the equation of a circle in the standard form where the center is at the origin \((0, 0)\) and the radius is \(\sqrt{7}\).
02

Identifying Key Features

To sketch the graph, identify the circle's center, which is at \((0, 0)\), and its radius, which is \(\sqrt{7}\), approximately 2.64. These will help position and size the circle on a coordinate plane.
03

Drawing the Circle

Using the center at \((0, 0)\), draw a circle with a radius reaching up to \(\sqrt{7}\) units from the center. This means that on each axis (x and y), the circle will touch the points approximately (0, 2.64), (0, -2.64), (2.64, 0), and (-2.64, 0).
04

Verifying the Graph

Double-check the graph by ensuring that any point \((x, y)\) on the circle satisfies the equation \(x^2 + y^2 = 7\). A known point is \((0, \sqrt{7})\), as substituting into the equation holds true.

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Key Concepts

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

Equation of a Circle
The equation of a circle is a vital concept in geometry. When an equation is given in the form \(x^2 + y^2 = r^2\), it represents a circle with a center at the point \((0, 0)\), known as the origin, on the coordinate plane. Here, \(r\) denotes the radius of the circle. It is important to note that \(r^2\) is the square of the radius.In the exercise provided, the equation \(x^2 + y^2 = 7\) is in this standard form. This equation defines a circle where:
  • The center is at the origin \((0, 0)\).
  • The squared radius \(r^2\) is 7, meaning the radius \(r\) is \(\sqrt{7}\).
Understanding the equation helps us quickly determine these characteristics about the circle. It tells us exactly how to draw and locate the circle on a coordinate plane.
Coordinate Plane
A coordinate plane is an essential tool for graphing equations in geometry and algebra. It is a two-dimensional surface defined by a horizontal line, the x-axis, and a vertical line, the y-axis. These axes intersect at the origin, point \((0, 0)\).On a coordinate plane:
  • Each point has a position expressed in terms of coordinates \((x, y)\).
  • The x-coordinate shows the horizontal position (left or right).
  • The y-coordinate shows the vertical position (up or down).
When graphing a circle, you'll plot it by using its center \((h, k)\) and radius \(r\). For the circle \(x^2 + y^2 = 7\), we pinpoint the center at \((0, 0)\) and measure \(\sqrt{7}\) from this point to all directions equally, creating a perfectly round shape.
Radius Calculation
The radius of a circle is the distance from the center of the circle to any point on its circumference. Calculating the radius is crucial for understanding and sketching the circle accurately.Given an equation \(x^2 + y^2 = r^2\), like \(x^2 + y^2 = 7\), we find the radius by taking the square root of the right-hand side, \(r^2\). Here's how to find the radius:
  • Identify \(r^2 = 7\) from the equation.
  • Calculate the square root to find \(r = \sqrt{7}\).
  • This results in \(r \approx 2.64\) when computed as a decimal.
The radius \(\sqrt{7}\) is approximately 2.64 units, meaning the circle extends this distance outwards from the center. By understanding this radius, you can accurately draw the circle within the coordinate plane.

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Most popular questions from this chapter

A spherical balloon is being inflated at a rate of \(\frac{9}{2} \pi \mathrm{ft}^{3} / \mathrm{min}\). Express its radius \(r\) as a function of time \(t\) (in minutes), assuming that \(r=0\) when \(t=0\).

Exer. 13-26: Sketch, on the same coordinate plane, the graphs of \(f\) for the given values of \(c\). (Make use of symmetry, shifting, stretching, compressing, or reflecting.) $$ f(x)=\sqrt{c x}-1 ; \quad c=-1, \frac{1}{9}, 4 $$

Exer. 53-60: Find a composite function form for \(y\). $$ y=\frac{\sqrt{x+4}-2}{\sqrt{x+4}+2} $$

Exer. 35-36: Solve the equation \((f \circ g)(x)=0\). $$ f(x)=x^{2}-2, \quad g(x)=x+3 $$

Exer. 21-34: Find (a) \((f \circ g)(x)\) and the domain of \(f \circ g\) and (b) \((g \circ f)(x)\) and the domain of \(g \circ f\). $$ f(x)=x^{2}, \quad g(x)=\frac{1}{x^{3}} $$
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