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Chapter 1: Problem 89

Show that the equation represents a circle, and find the center and radius of the circle. $$x^{2}+y^{2}-4 x+10 y+13=0$$

Short Answer

Expert verified
The circle has center (2, -5) and radius 4.

Step by step solution

01

Identify the General Equation of a Circle

The general equation of a circle is \( (x-h)^2 + (y-k)^2 = r^2 \), where \((h, k)\) is the center of the circle and \(r\) is the radius. Compare this with the given equation, \(x^2 + y^2 - 4x + 10y + 13 = 0\), to determine its form.
02

Rearrange the Given Equation

Reorder the terms: \(x^2 - 4x + y^2 + 10y = -13\). This step organizes the equation so that it is clearer for the process of completing the square.
03

Complete the Square for x

To complete the square for \(x\), take half of the coefficient of \(x\), which is -4. Half of -4 is -2. Then square -2 to get 4. Add and subtract 4 inside the equation: \((x^2 - 4x + 4) - 4\).
04

Complete the Square for y

To complete the square for \(y\), take half of the coefficient of \(y\), which is 10. Half of 10 is 5. Then square 5 to get 25. Add and subtract 25 inside the equation: \((y^2 + 10y + 25) - 25\).
05

Rearrange and Simplify

We now have \((x^2 - 4x + 4) + (y^2 + 10y + 25) = -13 + 4 + 25\). Simplify the completed squares and constant totals: \((x-2)^2 + (y+5)^2 = 16\).
06

Identify the Center and Radius

The equation \((x-2)^2 + (y+5)^2 = 16\) is now in standard form. Compare it to \((x-h)^2 + (y-k)^2 = r^2\) to find that the center \((h, k)\) is \((2, -5)\) and the radius \(r\) is \(\sqrt{16} = 4\).

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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 in algebra to transform quadratic equations into a perfect square trinomial. This is particularly useful when converting an equation to make it resemble a known standard form, such as that of a circle. Here鈥檚 how it works in the context of a circle's equation:

  • Firstly, group the like terms of the variables (\(x^2 - 4x\) and \(y^2 + 10y\) in this case).
  • For each variable group, take half of the linear coefficient (the number in front of \(x\) or \(y\)). For \(x\), half of \(-4\) is \(-2\), and for \(y\), half of \(10\) is \(5\).
  • Square these results to find the numbers to add and subtract within your equation, forming complete squares (\((-2)^2 = 4\) and \(5^2 = 25\)).
  • Your equation then features perfect square trinomials, looking like \((x-2)^2\) and \((y+5)^2\).
Completing the square allows you to rewrite any quadratic equation into a more understandable and workable format. This method is especially valuable for solving geometry-related algebra problems, like identifying the properties of a circle from its equation.
Center of a Circle
In the equation of a circle \((x - h)^2 + (y - k)^2 = r^2\), the center of the circle is represented by the coordinates \((h, k)\). This point is where the circle is perfectly balanced or centered.

To find the center from a standard circle equation, for example, \((x - 2)^2 + (y + 5)^2 = 16\), you need to:
  • Identify the terms inside each square. For \((x - 2)\), the \(-2\) corresponds to \(h = 2\).
  • Similarly, for \((y + 5)\), realize that \(+5\) is \(-(-5)\): here, \(k = -5\).
Thus, the center of this circle is \((2, -5)\).

Understanding the center's location helps visualize the circle's position in the coordinate plane.
Radius of a Circle
The radius of a circle is a line segment from the center of the circle to any point on its circumference. In the equation \((x - h)^2 + (y - k)^2 = r^2\), the radius is given by the term that remains when taking the square root of \(r^2\).

For example, consider \(r^2 = 16\) in the equation \((x-2)^2 + (y+5)^2 = 16\). The radius \(r\) is \(\sqrt{16} = 4\).
  • The radius indicates how far any point on the circle is from the center, \((h, k)\), which is \((2, -5)\)in our example.
  • Knowing the radius is crucial for understanding the size of the circle and graphing it accurately.
Visualizing the radius as a constant distance from the center makes it easier to sketch circles and solve geometric problems.

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Most popular questions from this chapter

The left-hand column in the table lists some common algebraic errors. In each case, give an example using numbers that show that the formula is not valid. An example of this type, which shows that a statement is false, is called a counterexample.$$\begin{array}{|c|c|} \hline \text { Algebraic error } & \text { Counterexample } \\ \hline \frac{1}{a}+\frac{1}{b}=\frac{1}{\sqrt{a}+b} & \frac{1}{2}+\frac{1}{2} \neq \frac{1}{2+2} \\ (a+b)^{2}=a^{2}+b^{2} & \\ \sqrt{a^{2}+b^{2}}=a+b & \\ \frac{a+b}{a}=b \\ \left(a^{3}+b^{3}\right)^{1 / 3}=a+b & \\ a^{m} / a^{n}=a^{m / n} & \\ a^{-1 / n}=\frac{1}{a^{n}} & \\ \hline \end{array}$$

A clothing manufacturer finds that the cost of producing \(x\) shirts is \(500+6 x+0.01 x^{2}\) dollars. (a) Explain why the average cost per shirt is given by the rational expression $$ A=\frac{500+6 x+0.01 x^{2}}{x} $$ (b) Complete the table by calculating the average cost per shirt for the given values of \(x .\) $$\begin{array}{|r|l|} \hline x & \text { Average cost } \\ \hline 10 & \\ 20 & \\ 50 & \\ 100 & \\ 200 & \\ 500 & \\ 1000 & \end{array}$$

Show that the equation represents a circle, and find the center and radius of the circle. $$x^{2}+y^{2}+\frac{1}{2} x+2 y+\frac{1}{16}=0$$

Factor the trinomial. $$x^{2}-6 x+5$$

Use a Special Factoring Formula to factor the expression. $$27 x^{3}+y^{3}$$
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