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Chapter 10: Problem 11

Find the center and radius of the circle with the given equation. Then graph the circle. $$ x^{2}+y^{2}+8 x-6 y=0 $$

Short Answer

Expert verified
Center: (-4, 3), Radius: 5

Step by step solution

01

Rearrange the equation

Let's start by rewriting the given circle equation to separate the terms involving the same variable: x^{2} + 8x + y^{2} - 6y = 0.
02

Complete the square for x-terms

Take the x-term part, which is x^2 + 8x. To complete the square, add and subtract (8/2)^2 = 16 within the equation: x^{2} + 8x + 16 - 16.
03

Complete the square for y-terms

For the y-term part, y^2 - 6y, add and subtract (6/2)^2 = 9 within the equation: y^{2} - 6y + 9 - 9.
04

Rewrite as perfect squares

Now, express the completed squares: (x + 4)^2 - 16 + (y - 3)^2 - 9.
05

Simplify the equation

Reorganize the equation: (x + 4)^2 + (y - 3)^2 = 16 + 9 = 25.
06

Identify the center and radius

The equation (x + 4)^2 + (y - 3)^2 = 25 is in the standard form of a circle equation, (x-h)^2 + (y-k)^2 = r^2, where (h, k) is the center and r is the radius.Thus, the center is (-4, 3) and the radius is \( \sqrt{25} = 5 \).

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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 method used to transform quadratic expressions into a more workable form, particularly a binomial squared. This technique is especially useful in deriving the standard form of a circle equation from its general form.
To complete the square for a quadratic expression such as \( x^2 + bx \), follow these steps:
  • Take half of the coefficient of the x-term, which is \( b \), and divide it by 2.
  • Square that result to get \( \left(\frac{b}{2}\right)^2 \).
  • Add and subtract \( \left(\frac{b}{2}\right)^2 \) to the expression. This doesn鈥檛 change the value but allows us to rewrite the expression as a perfect square trinomial.
For example: \( x^2 + 8x \) becomes \( x^2 + 8x + 16 - 16 \), which simplifies to \((x+4)^2 - 16\).
This manipulation makes it easier to graph circles and find crucial information like the center and radius.
Standard Form of a Circle Equation
The standard form of a circle's equation provides a simple way to describe a circle's size and location. The general format is:
\[ (x-h)^2 + (y-k)^2 = r^2 \]
In this equation,
  • \((h, k)\) represents the center of the circle.
  • \(r\) is the radius, which is the distance from the center to any point on the circle.
After completing the square on both the x and y terms in a general quadratic form, you can rearrange the equation to match this standard form.
For instance, if you have an equation like \( (x+4)^2 + (y-3)^2 = 25 \), you can immediately tell that the circle's center is at (-4, 3), and the radius is 5, since \( r^2 = 25 \) implies \( r = \sqrt{25} = 5 \).
This form is straightforward and immensely helpful for graphing and understanding the geometric properties of circles.
Graphing Circles
Graphing a circle involves plotting all the points that are a fixed distance, known as the radius, from a central point, the center, on a coordinate plane. Once your circle equation is in the standard form, graphing becomes a breeze.
Here鈥檚 how you can easily graph a circle:
  • Identify the center \((h, k)\) from the standard equation \((x-h)^2 + (y-k)^2 = r^2\). For example, the center in \((x+4)^2 + (y-3)^2 = 25\) is (-4, 3).
  • Determine the radius \(r\) by taking the square root of \(r^2\). With \(r^2 = 25\), the radius is 5.
  • Plot the center on the coordinate plane.
  • From the center, count the radius distance in all four cardinal directions (up, down, left, right) to establish the extent of the circle.
  • Draw a smooth curve connecting these boundary points in a symmetrical circle formation.
Graphing circles visually demonstrates their structure and helps in understanding their relationships with other geometric figures.

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Most popular questions from this chapter

Find the exact solution(s) of each system of equations. $$ \begin{array}{l} \frac{x^{2}}{36}-\frac{y^{2}}{4}=1 \\ x=y \end{array} $$

Write an equation of the hyperbola that satisfies each set of conditions. vertices \((6,-6)\) and \((0,-6),\) foci \((3 \pm \sqrt{13},-6)\)

Each equation is of the form \(A x^{2}+B x y+C y^{2}+D x+\) \(E y+F=0 .\) Identify the values of \(A, B,\) and \(C\). $$ -3 x^{2}+x y+2 y^{2}+4 x-7 y=0 $$

Solve each system of inequalities by graphing. $$ \begin{array}{l}{x^{2}+y^{2}<25} \\ {4 x^{2}-9 y^{2}<36}\end{array} $$

CHALLENGE A hyperbola with a horizontal transverse axis contains the point at \((4,3) .\) The equations of the asymptotes are \(y-x=1\) and \(y+x=5\) Write the equation for the hyperbola.
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