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Chapter 11: Problem 11

Find the center and radius of each circle. (Hint: In Exercises 15 and 16, divide each side by a common factor.) $$ x^{2}+y^{2}+4 x+6 y+9=0 $$

Short Answer

Expert verified
Center: (-2, -3), Radius: 2

Step by step solution

01

- Write the Standard Form of a Circle

Recall that the standard form of the 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.
02

- Group Terms and Move Constant

Group the \( x \) and \( y \) terms together and move the constant term to the other side of the equation: \[ x^2 + 4x + y^2 + 6y = -9 \]
03

- Complete the Square for \( x \)

To complete the square, take half of the coefficient of \( x \), which is 4, divide it by 2 to get 2, and square it to get 4. Add and subtract this value within the equation: \[ x^2 + 4x + 4 + y^2 + 6y = -9 + 4 \] This is equivalent to adding 4 to both sides of the equation.
04

- Complete the Square for \( y \)

To complete the square for \( y \), take half of the coefficient of \( y \), which is 6, divide it by 2 to get 3, and square it to get 9. Add and subtract this value within the equation: \[ x^2 + 4x + 4 + y^2 + 6y + 9 = -9 + 4 + 9 \] This is equivalent to adding 9 to both sides of the equation.
05

- Simplify and Write the Equation in Standard Form

Now simplify the equation: \[ (x + 2)^2 + (y + 3)^2 = 4 \] The equation is now in standard form.
06

- Identify the Center and Radius

From the standard form \( (x + 2)^2 + (y + 3)^2 = 4 \), the center of the circle is \( (-2, -3) \) and the radius is \( \sqrt{4} = 2 \).

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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 helpful method to transform a quadratic equation into a more workable form. We use it often to convert the general equation of a circle into its standard form.
It involves three main steps:

  • Take the coefficient of the linear term (like the coefficient of x or y), divide it by 2.
  • Square the result.
  • Add and subtract this square within the equation.
For example, in the given problem:
To complete the square for \( x^2 + 4x \), we took half of 4 (which is 2), and then squared it to get 4. So, we added and subtracted 4 within the equation. This helps us convert \( x^2 + 4x \) into \((x+2)^2 - 4\).
Similarly, we did this for \( y^2 + 6y\).
These steps help to easily convert the given equation into the standard form.
standard form of a circle
The standard form of a circle's equation is vital for identifying its key attributes easily. It looks like this:
\[ (x - h)^2 + (y - k)^2 = r^2 \]
In this form:
  • \((h, k)\) represents the center of the circle.
  • \(r\) represents the radius of the circle.
For understanding and solving problems involving circles, transforming the equation into this standard form is essential.
In our given problem, we transformed the general form \( x^2 + y^2 + 4x + 6y + 9 = 0 \) into the standard form \( (x + 2)^2 + (y + 3)^2 = 4 \). This helps us easily see the circle's center and radius.
circle center and radius
The center and radius are fundamental properties of a circle.
鈥� From the standard form \( (x - h)^2 + (y - k)^2 = r^2 \), we can quickly identify these.
鈥� The center is given by the coordinates \( (h, k) \), which are just the values opposite in sign to those inside the parentheses for x and y.
鈥� The radius is the square root of the value on the right side of the equation.
In our solution:
  • The center was \( (x + 2) \) and \( (y + 3) \), so the center is at \( (-2, -3) \).
  • For the radius, we had \( (x + 2)^2 + (y + 3)^2 = 4 \), which means the radius is \( \sqrt{4} = 2 \).
Understanding these elements is crucial for graphing and solving problems involving circles.

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Most popular questions from this chapter

use the fact that \(c^{2}=a^{2}-b^{2},\) where \(a^{2}>b^{2}\) The orbit of Venus around the sun (one of the foci) is an ellipse with equation $$ \frac{x^{2}}{5013}+\frac{y^{2}}{4970}=1 $$ where \(x\) and \(y\) are measured in millions of miles

Plot the points \((3,4),(-3,4),(3,-4),\) and \((-3,-4)\)

Graph each ellipse. $$ \frac{x^{2}}{9}+\frac{y^{2}}{25}=1 $$

Solve each system by the substitution method. $$ \begin{aligned} &x^{2}-x y+y^{2}=0\\\ &x-2 y=1 \end{aligned} $$

Solve each system by the substitution method. $$ \begin{aligned} &x y=4\\\ &3 x+2 y=-10 \end{aligned} $$
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