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Chapter 8: Problem 55

Decide whether each equation has a circle as its graph. If it does, give the center and radius. $$9 x^{2}+36 x+9 y^{2}=-32$$

Short Answer

Expert verified
It's a circle with center (-2, 0) and radius \(\frac{2}{3}\).

Step by step solution

01

Identify the general form of the circle equation

To determine if the equation represents a circle, we compare it to the standard form \((x-h)^2 + (y-k)^2 = r^2\). First, check if both \(x^2\) and \(y^2\) terms are present and have the same coefficient. We have \(9x^2\) and \(9y^2\), and both coefficients are equal, which indicates that the equation could represent a circle.
02

Simplify the given equation

The given equation is \(9x^2 + 36x + 9y^2 = -32\). First, factor 9 out from the terms involving \(x\) and \(y\): \[ 9(x^2 + 4x + y^2) = -32 \]. Then divide the entire equation by 9 to simplify: \[ x^2 + 4x + y^2 = -\frac{32}{9} \].
03

Complete the square for the \(x\) terms

To complete the square for the \(x\) terms, take half of the coefficient of \(x\), which is 4. Half of 4 is 2, and squared gives 4. Add and subtract 4 inside the equation: \[ x^2 + 4x + 4 - 4 + y^2 = -\frac{32}{9} \].
04

Rewrite the equation using perfect squares

The equation becomes \( (x+2)^2 - 4 + y^2 = -\frac{32}{9} \).We now group and simplify: \[ (x+2)^2 + y^2 = -\frac{32}{9} + 4 \].Simplifying the right side gives:\[ (x+2)^2 + y^2 = \frac{4}{9} \].
05

Identify the center and the radius

The equation \((x+2)^2 + y^2 = \frac{4}{9}\) is in the standard form of a circle \((x-h)^2 + (y-k)^2 = r^2\), with the center \((-2, 0)\) and the radius \(\sqrt{\frac{4}{9}} = \frac{2}{3}\). Therefore, the equation does indeed represent a circle.

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Key Concepts

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

Completing the Square
Completing the square is a useful skill, particularly when working with equations of circles. It helps to rewrite a quadratic equation into a form that makes it easier to interpret. Here's how it works:
  • Start with a quadratic expression, typically something like \(x^2 + bx\).
  • Take half of the coefficient of \(x\) (in this case, \(b\)), square it, and add it to the expression.
  • This action "completes" the square, allowing you to rewrite the expression as \((x + \frac{b}{2})^2\).
In the original exercise, the coefficient of \(x\) is 4. Hence, half of that is 2, and squaring 2 gives us 4. Adding and subtracting this value in the equation helps us create a perfect square trinomial. This process transforms the equation into a format that allows further simplification and conversion into the standard form of a circle equation.
Standard Form
The standard form of a circle's equation is crucial for understanding and graphing circles. The standard form is given by:
\((x-h)^2 + (y-k)^2 = r^2\)
  • Here, \((h, k)\) is the center of the circle.
  • \(r\) represents the radius.
In the exercise, we successfully transformed the equation into the standard form: \((x+2)^2 + y^2 = \frac{4}{9}\).
From this, it is evident that \(h = -2\) and \(k = 0\), which identifies the center. Additionally, the radius squared, \(r^2\), matches \(\frac{4}{9}\), from which we can determine \(r\). Moving from a general to standard form is key to discerning these attributes.
Center and Radius of a Circle
The stage is set once you've got the circle equation in standard form. Finding the center and radius becomes direct.
  • The center \((h, k)\) is taken directly from the equation \((x-h)^2 + (y-k)^2 = r^2\).
  • The radius \(r\) is simply the square root of the number on the right side of the equation.
In our specific equation \((x+2)^2 + y^2 = \frac{4}{9}\), you can see:
  • The center is \((-2, 0)\) because \(h\) is 2 with a sign change, and \(k\) is 0.
  • The radius \(r\) involves evaluating \(\sqrt{\frac{4}{9}}\), providing us the radius \(\frac{2}{3}\).
These parameters are not just numbers; they give the circle its identity, specifying exactly where it is positioned on the coordinate plane and how far it extends.

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Most popular questions from this chapter

CHECKING ANALYTIC SKILLS Graph each hyberbola by hand. Give the domain and range. Give the center in Exercises \(55-61 .\) Do not use a calculator. $$\frac{4 x^{2}}{9}-\frac{25 y^{2}}{16}=1$$

Graph the solution set of each system of inequalities by hand. $$\begin{aligned} &\frac{(x-1)^{2}}{9}+\frac{y^{2}}{4} \leq 1\\\ &\frac{x^{2}}{4}-\frac{(y+1)^{2}}{9} \geq 1 \end{aligned}$$

Write the equation in standard form for an ellipse centered at ( \(h, k\) ). Identify the center and vertices. $$4 x^{2}+8 x+y^{2}+2 y+1=0$$

Write an equation for each conic. Each parabola has vertex at the origin, and each ellipse or hyperbola is centered at the origin. $$y \text { -intercepts }(0, \pm 4) ; e=\frac{7}{3}$$

Use substitution to solve the nonlinear system of equations in three variables. Note that solutions are ondered triples. $$\begin{aligned} &x^{2}+y^{2}+z^{2}=4\\\ &x+y+z=2\\\ &x-y \quad=0 \end{aligned}$$
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