Chapter 10: Problem 59
Write each equation in standard form, if it is not already so, and graph it. The problems include equations that describe circles, parabolas, ellipses, and hyperbolas. $$ x^{2}+y^{2}+4 x-6 y-23=0 $$
Short Answer
Expert verified
Equation in standard form: \((x+2)^{2} + (y-3)^{2} = 36\); Center: (-2,3); Radius: 6.
Step by step solution
01
Identifying the Type of Equation
The given equation is \(x^{2}+y^{2}+4x-6y-23=0\). Since both \(x\) and \(y\) are squared and their coefficients are equal, the equation represents a circle.
02
Rewriting the Equation
We need to rewrite the equation in the standard form of a circle, which is \((x-h)^{2}+(y-k)^{2}=r^{2}\). Our goal will be to complete the square for both the \(x\) and \(y\) terms.
03
Completing the Square for x
For the \(x\) terms, the expression is \(x^{2}+4x\). Take half of the \(4\), which is \(2\), square it to get \(4\), and add \(4\) to both sides of the equation: \(x^{2} + 4x + 4\).
04
Completing the Square for y
For the \(y\) terms, the expression is \(y^{2}-6y\). Take half of \(-6\), which is \(-3\), square it to get \(9\), and add \(9\) to both sides of the equation: \(y^{2} - 6y + 9\).
05
Rewrite the Equation with Completed Squares
Adding these squares to the left side changes the original equation to: \((x+2)^{2} + (y-3)^{2} = 23 + 4 + 9\). Simplifying yields: \((x+2)^{2} + (y-3)^{2} = 36\).
06
Graphing the Circle
The equation \((x+2)^{2} + (y-3)^{2} = 36\) is in standard form. The center of the circle is \((-2, 3)\), and the radius is \(6\) (since \(r^{2}=36\), so \(r=6\)). Plot the center on the graph and draw a circle with radius 6.
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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 in algebra to transform a quadratic expression into a perfect square trinomial. This process is particularly useful when dealing with equations of circles, as it helps to rewrite those equations in standard form.
To complete the square for a term like \(x^2 + bx\), you follow these steps:
This method is essential for converting equations like \(x^2 + y^2 + 4x - 6y - 23 = 0\) into a form that is easily recognizable as a circle. By completing the square, you identify the transformations needed for graphing and interpreting the equation in terms of a circle’s position and size. Completing the square simplifies the equation and exposes features like the circle's center and radius.
To complete the square for a term like \(x^2 + bx\), you follow these steps:
- Identify the coefficient of \(x\), which is \(b\).
- Take half of \(b\), resulting in \(\frac{b}{2}\).
- Square this, \(\left(\frac{b}{2}\right)^2\), and add it to both sides of the equation to maintain equality.
This method is essential for converting equations like \(x^2 + y^2 + 4x - 6y - 23 = 0\) into a form that is easily recognizable as a circle. By completing the square, you identify the transformations needed for graphing and interpreting the equation in terms of a circle’s position and size. Completing the square simplifies the equation and exposes features like the circle's center and radius.
Standard Form of a Circle
The standard form of a circle's equation is given by \((x-h)^2 + (y-k)^2 = r^2\). In this equation, \((h, k)\) represents the center of the circle, and \(r\) is the radius. This formulation makes it straightforward to understand the circle's properties.
By completing the square on a general quadratic equation that includes both \(x^2\) and \(y^2\) terms with linear terms, you can identify \(h\), \(k\), and \(r\). For example, the equation \((x+2)^2 + (y-3)^2 = 36\) has:
By completing the square on a general quadratic equation that includes both \(x^2\) and \(y^2\) terms with linear terms, you can identify \(h\), \(k\), and \(r\). For example, the equation \((x+2)^2 + (y-3)^2 = 36\) has:
- Center at \((-2, 3)\)
- Radius of 6, since \(r^2 = 36\)
Graphing Circles
Graphing circles involves plotting a perfect round shape on a coordinate plane, based on its standard form equation \((x-h)^2 + (y-k)^2 = r^2\). The graph of a circle is defined by its center \((h, k)\) and radius \(r\).
Here are key steps to follow when graphing:
Here are key steps to follow when graphing:
- Identify the center \((h, k)\) from the equation.
- Determine the radius \(r\) by taking the square root of \(r^2\).
- Plot the center point on the graph.
- Use the radius to mark key points around the center in each direction – up, down, left, and right – on the coordinate plane.
- Draw a smooth curve connecting these points, ensuring it's equidistant from the center.