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Magazine Chapter 2: Problem 73
\(73-76\) Sketch the graph of the equation. $$ x^{2}+y^{2}+4 x-10 y=21 $$
Short Answer
Expert verified
The graph is a circle centered at (-2, 5) with a radius of \( 5\sqrt{2} \).
Step by step solution
01
Identify the Type of Equation
The given equation \( x^2 + y^2 + 4x - 10y = 21 \) is a quadratic equation in two variables and represents a conic section. We will attempt to rewrite this equation in the standard form of a circle by completing the square for both \( x \) and \( y \).
02
Rearrange the Equation
Group the \( x \) terms together and the \( y \) terms together: \( (x^2 + 4x) + (y^2 - 10y) = 21 \). This will help us to complete the square in both the \( x \) and \( y \) terms.
03
Complete the Square for x
To complete the square for the \( x \) terms, take half of the coefficient of \( x \), which is 4, divide it by 2 to get 2, and then square it to get 4. Add and subtract 4 inside the equation: \( (x^2 + 4x + 4 - 4) + (y^2 - 10y) = 21 \). This simplifies to \( (x + 2)^2 - 4 \).
04
Complete the Square for y
For the \( y \) terms, take half of -10, divide by 2 to get -5, and square it to get 25. Add and subtract 25: \( (x + 2)^2 - 4 + (y^2 - 10y + 25 - 25) = 21 \). This simplifies to \( (y - 5)^2 - 25 \).
05
Simplify the Equation
Insert the completed squares from Steps 3 and 4 back into the equation: \( (x + 2)^2 + (y - 5)^2 - 4 - 25 = 21 \). Combine constants: \( (x + 2)^2 + (y - 5)^2 = 50 \).
06
Identify the Characteristics of the Circle
The equation \( (x + 2)^2 + (y - 5)^2 = 50 \) is now in the standard form of a circle, \( (x - h)^2 + (y - k)^2 = r^2 \), where \( h = -2 \), \( k = 5 \), and \( r^2 = 50 \). Therefore, the center of the circle is (-2, 5) and the radius \( r = \sqrt{50} = 5\sqrt{2} \).
07
Sketch the Circle
Plot the center of the circle at (-2, 5) on a coordinate plane. Using the radius of \( 5\sqrt{2} \), draw a circle around this center. The circle should extend equally in all directions from the center (up, down, left, right) by the distance \( 5\sqrt{2} \).
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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 useful technique for solving quadratic equations and for rewriting quadratic expressions in a more manageable form. It's especially helpful when converting a general quadratic equation into one that's easier to interpret, such as when working with conic sections like circles.
To complete the square, follow these steps:
To complete the square, follow these steps:
- Identify the coefficient of the linear term. For example, in the term involving \( x \), look at the coefficient of \( x \).
- Divide this coefficient by 2 and then square the result. This gives you the number you need to add and subtract in the process.
- For an expression like \( x^2 + 4x \), you would divide 4 by 2, resulting in 2, then square it to get 4. Add and subtract this number within the expression to form a complete square.
- The expression \( x^2 + 4x \) becomes \( (x + 2)^2 - 4 \).
Equation of a Circle
The equation of a circle is a specific kind of quadratic equation that represents all the points (\( x, y \)) a fixed distance away from a central point (\( h, k \)) on a plane. The standard form for a circle's equation is:
\[(x-h)^2 + (y-k)^2 = r^2\]
\[(x-h)^2 + (y-k)^2 = r^2\]
- Here, \( h \) and \( k \) are the coordinates of the circle's center.
- \( r \) is the radius, or the fixed distance from the center to any point on the circle.
- In our given example, the equation \( (x + 2)^2 + (y - 5)^2 = 50 \) shows a circle centered at (-2, 5) with a radius of \( \sqrt{50} = 5\sqrt{2} \).
Quadratic Equation
A quadratic equation is any equation that can be rearranged in standard form as \( ax^2 + bx + c = 0 \), where \( a \), \( b \), and \( c \) are constants. This is one of the fundamental types of equations in algebra and is broadly applicable, especially in geometry and physics.
Quadratic equations can be solved via:
Quadratic equations can be solved via:
- Factoring, which involves rewriting the equation as a product of simpler binomials.
- Using the quadratic formula: \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). This formula provides the roots or solutions of the equation.
- Completing the square, which may also reveal special geometric forms like circles.
Coordinate Geometry
Coordinate geometry, also known as analytic geometry, is a branch of mathematics that connects algebra and geometry through graphs and equations. It's essential for visualizing mathematical concepts and interpreting equations geometrically, especially conic sections.
In coordinate geometry:
In coordinate geometry:
- Each point on a coordinate plane is identified by an ordered pair (\( x, y \)), describing its position relative to two perpendicular axes.
- Equations, like that of a circle, describe a set of points fulfilling certain algebraic conditions, offering a deep geometric insight.
- The focus can be on lines, circles, parabolas, ellipses, and hyperbolas. Each represents different relationships between algebraic terms.
- For example, the center and radius of the circle found in our exercise represent specific points and distances on this plane.
