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Chapter 1: Problem 13

In Exercises 11 through 14 , find the center and radius of each circle, and draw a sketch of the graph. $$ 3 x^{2}+3 y^{2}+4 y-7=0 $$

Short Answer

Expert verified
Center: (0, -2/3); Radius: 5/3

Step by step solution

01

Rewrite the equation

First, rewrite the given equation to group the x and y terms together. Divide the entire equation by 3 to simplify: \[ x^2 + y^2 + \frac{4}{3}y - \frac{7}{3} = 0 \]
02

Complete the square for the y terms

To complete the square for the y-terms, isolate the constant on one side: \[ x^2 + y^2 + \frac{4}{3}y = \frac{7}{3} \] Add and subtract the square of half the coefficient of y inside the equation: \[ y^2 + \frac{4}{3}y + \left(\frac{2}{3}\right)^2 = \frac{7}{3} + \left(\frac{2}{3}\right)^2 \] Thus, it becomes: \[ y^2 + \frac{4}{3}y + \frac{4}{9} = \frac{21}{9} + \frac{4}{9} \] Which simplifies to: \[ y^2 + \frac{4}{3}y + \frac{4}{9} = \frac{25}{9} \]
03

Write the equation in the standard form of a circle

Now express the completed square as a binomial square: \[ x^2 + \left( y + \frac{2}{3} \right)^2 = \frac{25}{9} \] The standard form of a circle equation is \( (x-h)^2 + (y-k)^2 = r^2 \). Here, \( h = 0 \), \( k = -\frac{2}{3} \), and \( r = \frac{5}{3} \). Hence, the equation of the circle is: \[ (x-0)^2 + \left( y + \frac{2}{3} \right)^2 = \left( \frac{5}{3} \right)^2 \]
04

Identify the center and radius

From the standard form, the center \((h,k)\) of the circle is \( (0, -\frac{2}{3}) \) and the radius \( r \) is \( \frac{5}{3} \).
05

Draw the graph

Sketch the circle with the identified center at \( (0, -\frac{2}{3}) \) and the radius of \( \frac{5}{3} \). Draw a circle starting from the center and extending \( \frac{5}{3} \) units away in all directions.

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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 a quadratic equation into a perfect square trinomial. This is helpful in finding the standard form of a circle's equation. For example, in the given problem:
Step 1 involves isolating the y-terms and completing the square for them. To complete the square, you:
  • Take the coefficient of the linear term (in this case, \frac{4}{3}\) and halve it.
  • Square that result (\(\frac{2}{3}\) becomes \(\frac{4}{9}\)).
  • Add and subtract this square inside the equation to maintain equality.
This allows the quadratic expression \(y^2 + \frac{4}{3} y + \frac{4}{9}\) to be written as a binomial square \( (y + \frac{2}{3})^2 \, making it easier to identify the center and radius of the circle.
Graphing Circles
Graphing a circle involves plotting all points that are at a fixed distance (the radius) from a central point (the center). From the completed square equation \( x^2 + ( y + \frac{2}{3})^2 = \frac{25}{9} \), we know that:
  • The center is at \(0, -\frac{2}{3}\)
  • The radius is \( \frac{5}{3} \)
When you graph this, remember:
  • Locate the center on the coordinate plane.
  • Measure the radius from the center in all directions to mark the circumference.
  • Draw a smooth curve through these points to create the circle.
Sketching the graph helps visually verify the solution and understand the geometry involved.
Radius and Center of a Circle
The radius and center of a circle are key components in its equation. Given the standard form:\( (x-h)^2 + (y-k)^2 = r^2 \),
  • The center \(h, k\) represents the coordinates from which every point on the circle is equidistant.
  • The radius \(r\) is the length of that fixed distance.
In our problem, we identified the standard form as \( (x-0)^2 + (y+\frac{2}{3})^2 = (\frac{5}{3})^2 \), Revealing:
  • The center \((h, k)\) is at \( (0, -\frac{2}{3}) \).
  • The radius \((r)\) is \( \frac{5}{3} \).
Understanding these components is crucial for graphing and solving circle-related problems. Always check the structure of the equation to correctly identify these elements.

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Most popular questions from this chapter

In Exercises 11 through 32 , find the solution set of the given inequality and illustrate the solution on the real number $$ 5 x+2>x-6 $$

In Exercises 1 through 10 , find the domain and range of the given function, and draw a sketch of the graph of the function. $$ f=\left\\{(x, y) \mid y=\sqrt{4-x^{2}}\right\\} $$

In Exercises 11 through 32 , find the solution set of the given inequality and illustrate the solution on the real number $$ \frac{2}{1-x} \leq 1 $$

Given \(G(x)=\sqrt{2 x^{2}+1}\), find: (a) \(G(-2)\) (b) \(G(0)\) (c) \(G\left(\frac{1}{b}\right)\) (d) \(G\left(\frac{4}{7}\right)\) (e) \(G\left(2 x^{2}-1\right)\) (f) \(\frac{G(x+h)-G(x)}{h}, h \neq 0\)

In Exercises 1 through 10, solve for \(x\). $$ |5-2 x|=11 $$
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