/*! 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 49 Decide whether each equation has... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 7: Problem 49

Decide whether each equation has a circle as its graph. If it does, give the center and radius. $$4 x^{2}+4 x+4 y^{2}-16 y-19=0$$

Short Answer

Expert verified
Yes, it's a circle with center \((-\frac{1}{2}, 2)\) and radius \(\frac{5}{2}\).

Step by step solution

01

Simplify the Equation to Identify Terms

Start with the given equation: \(4x^2 + 4x + 4y^2 - 16y - 19 = 0\). Notice that the terms involving \(x\) and \(y\) need to be separated for simplification. Divide the entire equation by 4 to make coefficients of \(x^2\) and \(y^2\) equal to 1: \(x^2 + x + y^2 - 4y - \frac{19}{4} = 0\).
02

Complete the Square for x-terms

To complete the square for the \(x\)-terms, take the \(x\) coefficient (1), divide by 2 to get \(\frac{1}{2}\), and then square it to get \(\frac{1}{4}\). Add and subtract \(\frac{1}{4}\) in the equation: \(x^2 + x + \frac{1}{4} - \frac{1}{4}+y^2 - 4y - \frac{19}{4} = 0\). This becomes \((x + \frac{1}{2})^2 - \frac{1}{4}\).
03

Complete the Square for y-terms

Now complete the square for the \(y\)-terms. Take the \(-4\), divide by 2 to get \(-2\), and square it to get \(4\). Add and subtract 4: \(y^2 - 4y + 4 - 4 - \frac{19}{4} = 0\). This simplifies to \((y - 2)^2 - 4\).
04

Rewrite the Equation in Standard Form

Replace the completed squares back in the equation: \((x + \frac{1}{2})^2 - \frac{1}{4} + (y - 2)^2 - 4 - \frac{19}{4} = 0\). Simplify further to \((x + \frac{1}{2})^2 + (y - 2)^2 = \frac{25}{4}\). Now the equation is in the form \((x-h)^2 + (y-k)^2 = r^2\).
05

Identify the Circle's Center and Radius

From \((x + \frac{1}{2})^2 + (y - 2)^2 = \frac{25}{4}\), identify the center \((h, k)\) as \((-\frac{1}{2}, 2)\) and the radius \(r = \sqrt{\frac{25}{4}} = \frac{5}{2}\). Since it can be rewritten in the form \((x-h)^2 + (y-k)^2 = r^2\), the equation describes a circle.

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
Completing the square is a handy technique in algebra. It's used to transform quadratic expressions into a perfect square trinomial, making them easier to work with.
The aim is to express an equation like \(x^2 + ax\) as a square; e.g., \((x+b)^2\). Here's a step-by-step rundown:
  • Start with the equation in the form \(x^2 + bx\).
  • Identify the coefficient of \(x\), which in our exercise is \(1\).
  • Divide this coefficient by 2. For our example, \(\frac{1}{2}\).
  • Square the result. Giving \(\left(\frac{1}{2}\right)^2 = \frac{1}{4}\).
  • Add and subtract this square inside the equation \(x^2 + x + \frac{1}{4} - \frac{1}{4}\).
  • The first three terms become a perfect square: \((x + \frac{1}{2})^2\).
By completing the square for both \(x\) and \(y\) terms, you can neatly reorganize the equation. This helps easily identify a circle鈥檚 equation by transforming it into a standardized form.
Center of a Circle
The center of a circle is a point that is equidistant from all points along the circle's perimeter. In mathematical terms, the center is represented by \((h, k)\) in a circle's standard equation:
\((x-h)^2 + (y-k)^2 = r^2\).Having a circle equation like \((x+\frac{1}{2})^2 + (y-2)^2 = \frac{25}{4}\) allows us to determine the center by identifying \((h, k)\).
Here's how:
  • Examine the coefficients added/subtracted inside the square terms. These values, with reversed signs, represent the circle's center.
  • In the equation \((x+\frac{1}{2})^2 + (y-2)^2 = \frac{25}{4}\):
    • \(h\) corresponds to \(-\frac{1}{2}\) because of the \((x+\frac{1}{2})^2\) term.
    • \(k\) is \(2\) from \((y-2)^2\).
  • Thus, the center is \((-\frac{1}{2}, 2)\).
With this understanding, locating the center simply involves reading the values reversed from what's inside the parentheses of your squares.
Radius of a Circle
The radius of a circle is the constant distance from its center to any point on its perimeter.
The radius is usually represented by \(r\) in the standard circle equation: \((x-h)^2 + (y-k)^2 = r^2\).In the context of your exercise, simplifying yielded the equation \((x+\frac{1}{2})^2 + (y-2)^2 = \frac{25}{4}\). This lets us find \(r\) by comparing to the standard form.
  • Note the right side of the equation is \(\frac{25}{4}\).
  • This value represents \(r^2\), the squared radius of the circle.
  • To find \(r\), simply take the square root: \(r = \sqrt{\frac{25}{4}}\).
  • Calculating gives \(r = \frac{5}{2}\).
Understanding how to determine the radius from an equation helps you fully grasp a circle's dimensions. The radius, lengthened or shortened, determines the size of the circle since it extends from the center to the boundary.

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

Design of a Radio Telescope The U.S. Naval Research Laboratory designed a giant radio telescope weighing 3450 tons. Its parabolic dish had a diameter of 300 feet, with a focal length (the distance from the focus to the parabolic surface) of 128.5 feet. Determine the maximum depth of the 300 -foot dish. (Source: Mar, J. and H. Liebowitz, Structure Technology for Large Radio and Radar Telescope Systems, MIT Press.)

Find an equation for each ellipse. Major axis of length 6 ; foci \((0,2)\) and \((0,-2)\)

Find an equation for each ellipse. Fociat \((-3,-3)\) and \((7,-3)\); the point ( \(2,1\) ) on ellipse

Find an equation for each ellipse. Center \((3,-2) ; a=5 ; c=3 ;\) major axis vertical

Graph each parabola by hand, and check using a graphing calculator. Give the vertex, axis, domain, and range. $$x=(y+1)^{2}$$
See all solutions

Recommended explanations on Math Textbooks

Probability and Statistics

Read Explanation

Applied Mathematics

Read Explanation

Geometry

Read Explanation

Decision Maths

Read Explanation

Logic and Functions

Read Explanation

Statistics

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.