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Chapter 12: Problem 33

Write the center-radius form of the circle with the given equation. Give the center and radius. $$3 x^{2}+3 y^{2}-12 x-24 y+12=0$$

Short Answer

Expert verified
Center: (2, 4) and Radius: 4

Step by step solution

01

- Factor out common factor

Start by factoring out the common factor of 3 from the given equation: \[3(x^2 + y^2 - 4x - 8y + 4) = 0\] Divide both sides of the equation by 3:\[x^2 + y^2 - 4x - 8y + 4 = 0\]
02

- Move constant term to the other side

Isolate the constant term by moving it to the other side of the equation: \[x^2 + y^2 - 4x - 8y = -4\]
03

- Complete the square for the x-terms

Complete the square for the x-terms: \[x^2 - 4x \] To do this, take half of the coefficient of x, square it, and add it to both sides of the equation. Half of -4 is -2, and (-2)^2 is 4. So add 4 to both sides: \[x^2 - 4x + 4 + y^2 - 8y = -4 + 4\]Which simplifies to:\[(x - 2)^2 + y^2 - 8y = 0\]
04

- Complete the square for the y-terms

Complete the square for the y-terms: \[y^2 - 8y\] To do this, take half of the coefficient of y, square it, and add it to both sides of the equation. Half of -8 is -4, and (-4)^2 is 16. So add 16 to both sides: \[(x - 2)^2 + y^2 - 8y + 16 = 0 + 16\]Which simplifies to: \[(x - 2)^2 + (y - 4)^2 = 16\]
05

- Write the center-radius form and determine the center and radius

The equation is now in the center-radius form: \[(x - 2)^2 + (y - 4)^2 = 16\] Identify the center \(h, k\) and the radius \(r\):The center is (2, 4) and the radius is \sqrt{16} = 4\.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

center-radius form
The center-radius form of a circle is a simple way to write the equation of a circle. This form is very useful in geometry and algebra because it clearly shows the center and the radius of the circle. The general equation for the center-radius form of a circle is:
\[ (x-h)^{2} + (y-k)^{2} = r^{2} \]
Here:
  • \
completing the square
Completing the square is an algebraic technique used to transform a quadratic equation into a perfect square trinomial, making the equation easier to solve or interpret. It is especially useful when converting the general form of a circle's equation to the center-radius form. To complete the square for a quadratic in the form \(ax^2 + bx + c\), follow these steps:
  • Step 1: Rearrange the equation to group the x-terms and y-terms on one side and the constant on the other side.
  • Step 2: For each group, take half of the linear coefficient (the number attached to x or y), square it, and add this square to both sides of the equation.
  • Step 3: Rewrite the quadratic expressions as perfect squares.
In our exercise, we applied these steps to the terms involving x and y separately, transforming the original equation into a form that clearly shows the circle's center and radius.
geometry
Understanding the circle's equation through geometry can help visualize its properties. A circle is defined as the set of all points in a plane that are the same distance (the radius) from a fixed point (the center).
In a coordinate plane, the equation of a circle reveals key geometric features:
  • The center, \((h, k)\), is the fixed point from which all points on the circle are equally distant.
  • The radius, \(r\), is the constant distance from the center to any point on the circle.
For example, in the exercise, the center is \((2, 4)\) and the radius is \(4\). Understanding these elements helps to graph the circle and apply geometric concepts to solve related problems.
radius calculation
Calculating the radius from the equation of a circle involves recognizing the squared term on the right-hand side of the center-radius form. The radius is derived from this term, as it represents the squared value of the radius.
For instance, in the equation \((x-2)^2 + (y-4)^2 = 16\), the number \(16\) is the square of the radius. To find the radius, take the square root of \(16\), which gives us:
  • \(r = \sqrt{16} = 4\)
This shows that the circle has a radius of \(4\). Identifying and calculating the radius accurately is crucial for applications in both algebra and geometry, ensuring that you can fully describe and work with the circle's properties.

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Most popular questions from this chapter

The Schwab Company designs and sells two types of rings: the VIP and the SST. The company can produce up to 24 rings each day using up to 60 total hours of labor. It takes \(3 \mathrm{hr}\) to make one VIP ring and \(2 \mathrm{hr}\) to make one SST ring. The profit on one VIP ring is \(\$ 30,\) and the profit on one SST ring is \(\$ 40 .\) How many of each type of ring should be made daily to maximize profit? What is the maximum profit?

Solve each system using the substitution method. \(x^{2}-3 x+y^{2}=4\) \(2 x-y=3\)

Which one of the following is a description of the graph of the solution set of the following system? \(x^{2}+y^{2}<25\) \(y>-2\) A. All points outside the circle \(x^{2}+y^{2}=25\) and above the line \(y=-2\) B. All points outside the circle \(x^{2}+y^{2}=25\) and below the line \(y=-2\) C. All points inside the circle \(x^{2}+y^{2}=25\) and above the line \(y=-2\) D. All points inside the circle \(x^{2}+y^{2}=25\) and below the line \(y=-2\)

Write the center-radius form of the circle with the given equation. Give the center and radius, and graph the circle. $$x^{2}+y^{2}-4 x-6 y+9=0$$

A certain manufacturing process requires that oil refineries manufacture at least 2 gal of gasoline for each gallon of fuel oil. To meet the winter demand for fuel oil, at least 3 million gal per day must be produced. The demand for gasoline is no more than 6.4 million gal per day. If the price of gasoline is \(\$ 2.90\) per gal and the price of fuel oil is \(\$ 2.50\) per gal, how much of each should be produced to maximize revenue? What is the maximum revenue?
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