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Chapter 0: Problem 57

Find the center and radius of each circle. $$x^{2}+y^{2}-6 x-4 y+1=0$$

Short Answer

Expert verified
The center of the circle is (3, 2), and the radius is \(2\sqrt{3}\).

Step by step solution

01

Group Terms

Rewrite the given equation by grouping the x terms together and the y terms together. The given equation is: \( x^2 + y^2 - 6x - 4y + 1 = 0 \). Group the x terms as \( x^2 - 6x \) and the y terms as \( y^2 - 4y \).
02

Complete the Square for x

To complete the square for the x terms, take the coefficient of x (which is -6), halve it to get -3, then square it to get 9. Add and subtract 9 within the x terms: \( x^2 - 6x + 9 - 9 \). This becomes \( (x-3)^2 - 9 \).
03

Complete the Square for y

To complete the square for the y terms, take the coefficient of y (which is -4), halve it to get -2, then square it to get 4. Add and subtract 4 within the y terms: \( y^2 - 4y + 4 - 4 \). This becomes \( (y-2)^2 - 4 \).
04

Rearrange the Equation

Insert the completed squares back into the original equation: \((x-3)^2 - 9 + (y-2)^2 - 4 + 1 = 0\). Simplify it to: \((x-3)^2 + (y-2)^2 - 12 = 0\).
05

Solve for the Center and Radius

Move the constant to the right side of the equation: \((x-3)^2 + (y-2)^2 = 12 \). The standard form of a circle is \((x-h)^2 + (y-k)^2 = r^2\), where \((h, k)\) is the center and \(r\) is the radius. Here, the center \((h, k)\) is \((3, 2)\) and \((r^2 = 12)\), thus \(r = \sqrt{12} = 2\sqrt{3}\).

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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 technique used in algebra to solve quadratic equations or to transform them into a different form. This method allows us to rewrite a quadratic equation into a perfect square trinomial, which is incredibly helpful when working with the equation of a circle.

For the equation of a circle given in the problem, the goal of completing the square is to gather and rearrange terms so that each variable forms a perfect binomial square.
  • Start by focusing on the x terms. In our case: \( x^2 - 6x \). Take the linear term鈥檚 coefficient (in this example, \(-6\)), divide it by 2, giving \(-3\), and then square it, resulting in 9.
  • Similarly, with y terms \( y^2 - 4y \): halve the \(-4\) to obtain \(-2\), and square it to get 4.
These calculations transform the separate terms into squares like \((x-3)^2\) and \((y-2)^2\), simplifying the equation and making other characteristics of the circle clearer.
Center of a Circle
The equation of a circle in its standard form is \[ (x-h)^2 + (y-k)^2 = r^2 \] where \((h, k)\) is the center of the circle. From this form, you can directly identify the circle's center by examining the constants tied to the \(x\) and \(y\) terms.

In our worked example, after completing the square, the equation transforms to \[ (x-3)^2 + (y-2)^2 = 12 \].
  • Here, \(h = 3\) and \(k = 2\), indicating the center of the circle is at the point \((3, 2)\).
This straightforward method shows just how essential completing the square is in revealing the center from the general form of a circle's equation.
Radius of a Circle
Understanding the radius of a circle involves connecting the circle's equation to the geometric definition of a radius, which is the distance from the circle's center to any point on its perimeter. In standard form, the radius \(r\) is extracted from the formula \[ r = \sqrt{r^2} \].

In our equation, following the simplification \[ (x-3)^2 + (y-2)^2 = 12 \],
  • the value on the right is \(r^2 = 12\).
  • By calculating \(r = \sqrt{12}\), we determine the radius to be \(2\sqrt{3}\),indicating how all values fit together to describe the circle's size and position.
Recognizing that the square root is involved underlines how specific transformations and calculations lead to the full comprehension of the circle's dimensions.

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Most popular questions from this chapter

Refer to the following: From March 2000 to March 2008, data for retail gasoline price in dollars per gallon are given in the table below. (These data are from Energy Information Administration, Official Energy Statistics from the U.S. Government at http://tonto.eia.doe.gov/oog/info/gdu/gaspump.html.) Use the calculator [STAT] [EDIT] command to enter the table below with \(L_{1}\) as the year (x 1 for year 2000) and \(L_{2}\) as the gasoline price in dollars per gallon. $$\begin{array}{|l|c|c|c|c|c|c|c|c|c|}\hline \text { March of Each vear } & 2000 & 2001 & 2002 & 2003 & 2004 & 2005 & 2006 & 2007 & 2008 \\\\\hline \text { Rerale casoune Price } \$ \text { PER cautuon } & 1.517 & 1.409 & 1.249 & 1.693 & 1.736 & 2.079 & 2.425 & 2.563 & 3.244 \\\\\hline\end{array}$$ a. Use the calculator commands [STAT] [LinReg]to model the data using the least squares regression. Find the equation of the least squares regression line using \(x\) as the year \((x=1 \text { for year } 2000)\) and \(y\) as the gasoline price in dollars per gallon. Round all answers to three decimal places. b. Use the equation to determine the gasoline price in March \(2006 .\) Round all answers to three decimal places. Is the answer close to the actual price? c. Use the equation to find the gasoline price in March \(2009 .\) Round all answers to three decimal places.

Graph the function represented by each side of the equation in the same viewing rectangle and solve for \(x\) $$3(x+2)-5 x=3 x-4$$

Show that two lines with equal slopes and different \(y\) -intercepts have no point in common. Hint: Let \(y_{1}=m x+b_{1}\) and \(y_{2}=m x+b_{2}\) with \(b_{1} \neq b_{2} .\) What equation must be true for there to be a point of intersection? Show that this leads to a contradiction.

Solve each formula for the specified variable. \(A=l w\) for \(w\)

Explain the mistake that is made. Find the slope of the line that passes through the points (-2,3) and (4,1) Solution: Write the slope formula. \(\quad m=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\) Substitute (-2,3) and \((4,1), \quad m=\frac{1-3}{-2-4}=\frac{-2}{-6}=\frac{1}{3}\) This is incorrect. What mistake was made?
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