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Chapter 13: Problem 26

The graph of each equation is a circle. Find the center and the radius, and then graph the circle. See Examples 5 through 7. $$ x^{2}+y^{2}+6 x-4 y=3 $$

Short Answer

Expert verified
Center: (-3, 2), Radius: 4.

Step by step solution

01

Rearrange the Equation

The given equation is \( x^2 + y^2 + 6x - 4y = 3 \). First, we need to rearrange the terms so that all terms involving \( x \) are together, and all terms involving \( y \) are together. Thus, rewrite the equation as \( x^2 + 6x + y^2 - 4y = 3 \).
02

Complete the Square for the x-Terms

To complete the square for the \( x \) terms, take the coefficient of \( x \), which is 6, divide it by 2 to get 3, and then square it to get 9. Add and subtract 9 within the equation, resulting in \( x^2 + 6x + 9 - 9 + y^2 - 4y = 3 \).
03

Complete the Square for the y-Terms

To complete the square for the \( y \) terms, take the coefficient of \( y \), which is -4, divide it by 2 to get -2, and then square it to get 4. Add and subtract 4 within the equation, resulting in \( x^2 + 6x + 9 + y^2 - 4y + 4 = 3 + 9 + 4 \).
04

Write the Equation in Standard Form

The equation now is \( (x^2 + 6x + 9) + (y^2 - 4y + 4) = 16 \), which can be rewritten using the completed squares: \((x+3)^2 + (y-2)^2 = 16 \).
05

Identify the Center and Radius

The circle's equation is in the standard form \((x-h)^2 + (y-k)^2 = r^2 \). By comparing, we find that the center is \((-3, 2)\), and the radius is \( \sqrt{16} = 4 \).
06

Graph the Circle

Plot the center of the circle at \((-3, 2)\) on the Cartesian plane. From this point, draw a circle with a radius of 4 units. Ensure the circle is symmetrically around the center.

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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 helpful algebraic technique used to transform certain quadratic expressions into a form that is easier to work with. In the context of circles, this technique aids in rewriting the circle equation into its standard form. Here's a simple way to understand it:

  • Take the term involving the linear variable (for example, if you have \(x^2 + 6x\), focus on the \(6x\)).
  • Divide the coefficient of the variable by 2, and then square it. Here, \(6/2 = 3\) and \(3^2 = 9\). Add and subtract this square within the expression.
  • This allows you to restructure the quadratic and linear terms into a perfect square trinomial, \((x+3)^2\) in our example after completing the square.
You perform similar operations for the \(y\) variable. This process allows you to convert the general circle equation into one where the center and radius are easy to visualize.
Standard Form of a Circle
The standard form of a circle's equation is an essential format for understanding the geometric properties of a circle. It is expressed as \((x-h)^2 + (y-k)^2 = r^2\), where:

  • \( (h, k) \) is the center of the circle.
  • \( r \) is the radius.
By rewriting the equation using completing the square, you can see the center and radius clearly. For the equation \((x+3)^2 + (y-2)^2 = 16\):

  • The center \((h, k)\) is \((-3, 2)\).
  • The radius \(r\) is \(4\), once you take the square root of 16.
This form provides a straightforward way to understand and graph the circle accurately on a coordinate plane.
Graphing Circles
Graphing a circle is a straightforward process once you have its equation in standard form. Here's how you can do it effectively:

  • Identify the center \((h, k)\) from the equation \((x-h)^2 + (y-k)^2 = r^2\). This is the point around which your circle is centered. In our case, it's \((-3, 2)\).
  • Determine the circle's radius \(r\). It's the square root of the number on the right side of the equation. Here it's \(4\), which means the circle extends 4 units from the center in all directions.
When graphing:

  • Place the center at \((-3, 2)\) on a Cartesian plane.
  • From the center, mark points 4 units away, horizontally and vertically. These define the edges of your circle.
  • Draw a smooth round curve connecting these points. This represents all points equidistant from the center, forming the circle.
By following these steps, you visualize the circle effectively, ensuring it accurately represents the equation.

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