Chapter 10: Problem 50
Write each equation in standard form, if it is not already so, and graph it. The problems include equations that describe circles, parabolas, and ellipses. \(x^{2}+y^{2}+4 x+6 y+9=0\)
Short Answer
Expert verified
The circle has a center at (-2, -3) and a radius of 2.
Step by step solution
01
Recognize the Type of Equation
The given equation is of the form \(x^2 + y^2 + Ax + By + C = 0\), which is a standard form for the equation of a circle.
02
Rearrange the Terms
First, rewrite the equation:\[x^2 + 4x + y^2 + 6y + 9 = 0\] Group the \(x\) terms together and the \(y\) terms together.
03
Complete the Square for x
To complete the square for the \(x\) terms, take the coefficient of \(x\) (which is 4), divide by 2, and square it:\(\left(\frac{4}{2}\right)^2 = 4\).Add and subtract this inside the equation:\[x^2 + 4x + 4 - 4 + y^2 + 6y + 9 = 0\]\[ \implies (x + 2)^2 + y^2 + 6y + 5 = 0\]
04
Complete the Square for y
For the \(y\) terms, take the coefficient of \(y\) (which is 6), divide by 2, and square it:\(\left(\frac{6}{2}\right)^2 = 9\). Similarly, add and subtract this value:\[(x + 2)^2 + (y + 3)^2 - 9 + 5 = 0\] \[ \implies (x + 2)^2 + (y + 3)^2 - 4 = 0\]
05
Simplify to Standard Circle Form
Rearrange the equation by moving the constant to the other side:\[(x + 2)^2 + (y + 3)^2 = 4\]This is now in the standard form of a circle, \((x - h)^2 + (y - k)^2 = r^2\) where \(h = -2\), \(k = -3\), and \(r = 2\).
06
Graph the Circle
Plot the center of the circle at \((-2, -3)\). The radius is 2, so sketch a circle with a radius of 2 units around the center.
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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 key mathematical technique used to rewrite quadratic expressions. It's especially useful when working with equations of circles, as it helps to transform them into a more manageable form. Here's how completing the square operates step-by-step:
- Start with the equation parts that you need to square: in a circle equation, these are usually the terms with both the variable and its coefficient, such as \(4x\) and \(6y\) in our example.
- Take the coefficient of the variable, divide it by 2, and then square it. For \(4x\), you get \(4/2 = 2\), and squaring \(2\) provides \(4\). Similarly, for \(6y\), you compute \((6/2)^2 = 9\).
- Add and subtract the square of the halved coefficient within the expression. This step ensures the balance of the equation while forming a perfect square trinomial.
- Rewrite the expression using the perfect square trinomial, such as turning \(x^2 + 4x + 4\) into \((x + 2)^2\).
Standard Form of a Circle
The standard form of a circle's equation is a specific arrangement that makes the circle's characteristics easy to see. For a circle, the equation in standard form looks like this: \((x - h)^2 + (y - k)^2 = r^2\). Here,
- \(h\) and \(k\) represent the coordinates of the circle's center.
- \(r\) is the circle's radius.
Graphing Circles
Graphing a circle starts with simple steps once you have the equation in standard form. Here's a straightforward approach to graph a circle:
- Identify the circle's center from the equation. For example, from \((x + 2)^2 + (y + 3)^2 = 4\), you pinpoint \((-2, -3)\) as the center.
- Determine the radius. Since \(4 = r^2\), calculate the radius \(r = \sqrt{4} = 2\).
- Plot the center on the coordinate plane.
- Using the radius, count 2 units away from the center in all directions—up, down, left, and right—marking these points will define the circle's edge.
- Connect these markers smoothly and evenly, maintaining a circle shape around the center.