Chapter 13: Problem 53
Sketch the graph of each equation. If the graph is a parabola, find irs vertex. If the graph is a circle, find its center and radius. $$x^{2}+y^{2}+2 x+12 y-12=0$$
Short Answer
Expert verified
The graph is a circle with center (-1, -6) and radius 7.
Step by step solution
01
Identify the Type of Graph
The given equation is \(x^2 + y^2 + 2x + 12y - 12 = 0\). Since both \(x^2\) and \(y^2\) are present and both have a positive coefficient, the equation represents a circle.
02
Rearrange the Equation
Rewrite the equation by grouping the \(x\) and \(y\) terms together: \(x^2 + 2x + y^2 + 12y = 12\).
03
Complete the Square for x
To complete the square for \(x^2 + 2x\), take the coefficient of \(x\), divide by 2, and square it. Add and subtract this inside the equation:\[x^2 + 2x = (x + 1)^2 - 1\]
04
Complete the Square for y
To complete the square for \(y^2 + 12y\), take the coefficient of \(y\), divide by 2, and square it. Add and subtract this inside the equation:\[y^2 + 12y = (y + 6)^2 - 36\]
05
Substitute Completed Squares into the Equation
Substitute the completed squares back into the rearranged equation:\[(x + 1)^2 - 1 + (y + 6)^2 - 36 = 12\]Simplify to get:\[(x+1)^2 + (y+6)^2 = 49\]
06
Determine Center and Radius
The equation \((x+1)^2 + (y+6)^2 = 49\) can be compared with the standard form \((x-h)^2 + (y-k)^2 = r^2\), where \((h, k)\) is the center and \(r\) is the radius.So, the center is \((-1, -6)\) and the radius is \(\sqrt{49} = 7\).
07
Sketch the Graph
Plot the circle with center \((-1, -6)\) and radius 7 on the coordinate plane. The circle will have points on it that are exactly 7 units away from the center, 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 quadratic equations into a more manageable form, especially when working with circle equations. This technique helps in rewriting the equation into a format that is easier to graph or interpret geometrically.
When dealing with quadratics such as
When dealing with quadratics such as
- \(x^2 + 2x\),
- Take the coefficient of the linear term (here, it's 2 for the expression \(2x\)).
- Divide it by 2, resulting in 1.
- Square this number to provide the complete square term, resulting in 1.
- \((x + 1)^2 - 1\).
- \(y^2 + 12y\)
- \((y + 6)^2 - 36\).
Graphing Circles
Graphing circles involves transforming them into a standard form, making visualizing these geometric shapes easier and more accurate on a coordinate plane.
The standard form of a circle's equation is
The standard form of a circle's equation is
- \((x - h)^2 + (y - k)^2 = r^2\).
- \((h, k)\) represents the circle's center.
- \(r\) is the radius.
- such as \((-1, -6)\) from the example equation,
- (\(r = 7\) in this case)
Coordinate Geometry
Coordinate geometry provides a rich framework for analyzing and understanding shapes within the two-dimensional Cartesian coordinate system. It's particularly useful for describing geometric shapes like circles in terms of algebraic equations.
When dealing with coordinate geometry, circles are often expressed using a particular equation, namely
When dealing with coordinate geometry, circles are often expressed using a particular equation, namely
- \((x - h)^2 + (y - k)^2 = r^2\).
- the circle's center \((h, k)\), and
- the radius \(r\).