/*! 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 42 Rewrite each equation in the sta... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 12: Problem 42

Rewrite each equation in the standard form for the equation of a circle, and identify its center and radius. $$x^{2}-3 x+y^{2}=0$$

Short Answer

Expert verified
Center: (\frac{3}{2}, 0). Radius: \frac{3}{2}.

Step by step solution

01

Rewrite the equation and group terms

Start by rewriting the given equation and grouping the terms involving the same variable together:\[x^2 - 3x + y^2 = 0\]To complete the square, we'll isolate the terms involving \(x\) and \(y\) separately.
02

Complete the square for the \(x\)-terms

To complete the square for the \(x\)-terms, take half of the linear coefficient of \(x\), square it, and add and subtract this value inside the equation:\[x^2 - 3x\text{ }\to\text{ } x^2 - 3x + \frac{9}{4} - \frac{9}{4} \ = (x - \frac{3}{2})^2 - \frac{9}{4}\]
03

Combine the completed square with the original equation

Now, substitute the completed square back into the equation, combined with the original \(y^2\) term:\[(x - \frac{3}{2})^2 - \frac{9}{4} + y^2 = 0\]
04

Simplify and isolate the completed square term

Next, simplify and isolate the completed square term and the constant term by combining them on the same side of the equation:\[(x - \frac{3}{2})^2 + y^2 = \frac{9}{4}\]
05

Rewrite in standard form and identify center and radius

The equation \[(x - \frac{3}{2})^2 + y^2 = \frac{9}{4}\] is in standard form for a circle: \((x - h)^2 + (y - k)^2 = r^2\).Identify the center \( (h, k) \): \( (\frac{3}{2}, 0) \)Identify the radius \( r \): \( \frac{3}{2} \)

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

completing the square
To understand completing the square, think of it as a method to transform a quadratic expression into a perfect square trinomial. This technique helps in rewriting the equation of a circle in standard form. Consider the quadratic expression from the given problem:
Let's take the term involving 'x', which is \( x^2 - 3x \).
To complete the square, follow these steps:
  • Take half of the linear coefficient (the coefficient of x) which is -3. Half of -3 is -\(\frac{3}{2}\).
  • Square this result: \(( -\frac{3}{2} )^2 = \frac{9}{4}\).
  • Add and subtract this squared value inside the equation: \(x^2 - 3x + \frac{9}{4} - \frac{9}{4}\). This turns our expression into: \((x - \frac{3}{2})^2 - \frac{9}{4}\).
By doing this, you have transformed \( x^2 - 3x \) into a perfect square trinomial. Now, you can easily work with it in the equation.
standard form
The standard form of the equation of a circle presents it in an easily recognizable format. The general standard form is: \[ (x - h)^2 + (y - k)^2 = r^2 \]

Here:
  • \( (x - h) \) represents the horizontal component from the center of the circle.
  • \( (y - k) \) represents the vertical component from the center of the circle.
  • \( r \) is the radius of the circle.
Let's look at how the given equation transforms into standard form: Given equation: \[ x^2 - 3x + y^2 = 0 \]

  1. First, complete the square for the x-terms, resulting in \( (x - \frac{3}{2})^2 - \frac{9}{4} \).
  2. Next, substitute this completed square back into the equation along with \( y^2 \): \[ (x - \frac{3}{2})^2 - \frac{9}{4} + y^2 = 0 \]
  3. Combine and simplify the constants on one side: \[ (x - \frac{3}{2})^2 + y^2 = \frac{9}{4} \]
This final form is the standard form of the circle's equation, making it easy to identify the center and radius.
center and radius
The center and radius are key features of a circle. In standard form, the equation of a circle can be written as: \[ (x - h)^2 + (y - k)^2 = r^2 \]

From this equation:
  • The center of the circle is at coordinates (h, k).
  • The radius is \( r \).
Using the transformed equation from the previous section: \[ (x - \frac{3}{2})^2 + y^2 = \frac{9}{4} \]

  • We can see that the center (h, k) is derived from the expression inside the parentheses: \(x - \frac{3}{2}\) and \(y \). Thus, the center is \( (\frac{3}{2}, 0) \).
  • To find the radius, take the square root of the constant term on the right side of the equation: \( \frac{9}{4} \). The radius \( r \) is therefore \( \frac{3}{2} \).
Knowing the center and radius helps visualize and plot the circle accurately in a coordinate system.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Graph both equations of each system on the same coordinate axes. Use elimination of variables to find all points of intersection. $$ \begin{aligned} &\frac{x^{2}}{4}+\frac{y^{2}}{16}=1\\\ &x^{2}+y^{2}=1 \end{aligned} $$

Graph each hyperbola and write the equations of its asymptotes. $$ y^{2}-x^{2}=1 $$

Sketch the graph of each parabola. $$ x=-2(y-1)^{2}+3 $$

$$\text {Graph each equation.}$$ $$y=\sqrt{1-x^{2}}$$

Sketch the graph of each ellipse. $$ (x-2)^{2}+\frac{(y+1)^{2}}{36}=1 $$
See all solutions

Recommended explanations on Math Textbooks

Probability and Statistics

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Discrete Mathematics

Read Explanation

Decision Maths

Read Explanation

Mechanics Maths

Read Explanation

Calculus

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.