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Chapter 10: Problem 7

Find the center and radius of the circle described in the given equation. $$x^{2}+2 x+y^{2}=4$$

Short Answer

Expert verified
Center: (-1, 0), Radius: \(\sqrt{5}\).

Step by step solution

01

Organize the Equation

Start with the given equation of the circle: \[ x^2 + 2x + y^2 = 4 \]Our goal is to transform this equation into the standard form of a circle, which is \((x-h)^2 + (y-k)^2 = r^2\), where \((h, k)\) is the center and \(r\) is the radius.
02

Complete the Square for x-term

To complete the square for the \(x\)-terms, take the coefficient of \(x\), which is 2, divide by 2 and square it. So, \(\left(\frac{2}{2}\right)^2 = 1\).Add and subtract 1 inside the equation to complete the square:\[ x^2 + 2x + 1 - 1 + y^2 = 4 \]
03

Reorganize the Equation

Rewrite the equation as:\[ (x+1)^2 - 1 + y^2 = 4 \]Now add 1 on both sides of the equation to balance it:\[ (x+1)^2 + y^2 = 5 \]
04

Identify the Center and Radius

Compare the equation \((x+1)^2 + y^2 = 5\) with the standard form \((x-h)^2 + (y-k)^2 = r^2\).- The center \((h, k)\) is \((-1, 0)\).- The radius \(r\) is \(\sqrt{5}\).

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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 mathematical technique to transform a quadratic equation into a perfect square trinomial. This concept is vital in transforming the general form of a circle's equation to its standard form, making it more informative. If you have an equation like \(x^2 + 2x\), you can complete the square by:
  • Identifying the coefficient of \(x\), which is 2 in this case.
  • Halving the coefficient: \(\frac{2}{2} = 1\).
  • Squaring the result: \(1^2 = 1\).
You then add and subtract this squared term (1) within the equation to form a perfect square trinomial, \((x+1)^2\). This step simplifies the equation, making subsequent steps to find the circle's standard form and other properties much easier. It is a small change with a powerful impact in making the equation neat and interpretable.
Standard Form of a Circle
The standard form of a circle's equation is derived to present its key features clearly. The equation \((x-h)^2 + (y-k)^2 = r^2\) reveals the circle's center and radius at a glance:
  • \( (h, k) \) - the coordinates representing the center of the circle.
  • \( r \) - the radius of the circle.
Converting a general or expanded form equation to this standard form can involve techniques such as completing the square. This step-by-step simplification ensures the equation is concise and informative.In our example, after completing the square, the equation was reorganized neatly to \((x+1)^2 + y^2 = 5\).This format makes it clear that the circle is centered at \((-1, 0)\) with a radius of \(\sqrt{5}\).
Center and Radius of a Circle
The main goal of converting a circle's equation into standard form is to easily identify its center and radius. These components - the center \((h, k)\) and radius \(r\) - are pivotal for defining a circle's geometry:
  • The center \((h, k)\) determines the position of the circle in the coordinate plane.
  • The radius \(r\) describes the distance from the center to any point on the circle's circumference.
From the standard form equation \((x-h)^2 + (y-k)^2 = r^2\), you can directly read off these values. For example, in the equation \((x+1)^2 + y^2 = 5\), comparing to the form \((x-h)^2 + (y-k)^2 = r^2\) tells us:- The center is at \((-1, 0)\), as \(h = -1\) and \(k = 0\).- The radius is \(\sqrt{5}\), since \(r^2 = 5\) so \(r = \sqrt{5}\).Understanding and extracting these elements from the equation is crucial for any geometric analysis of circles.

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