Problem 39
Question
Write the standard form of the equation of the circle with the given center and radius. Center \((-4,0), r=10\)
Step-by-Step Solution
Verified Answer
The standard form of the equation of the circle with the given center \(-4,0\) and radius \(10\) is \((x+4)^2 + y^2 = 100\).
1Step 1: Identify the Center and Radius
In the question, the center (a,b) of the circle is given as \(-4,0\) and the radius \(r\) as \(10\).
2Step 2: Substitute into the Standard Equation
Substitute \(-4\) for \(a\), \(0\) for \(b\), and \(10\) for \(r\) in \((x-a)^2+(y-b)^2 = r^2\). The substituted equation is \((x--4)^2+(y-0)^2 = 10^2\).
3Step 3: Simplify the Substituted Equation
Simplify the substituted equation into a more standard form. The calculation becomes \((x+4)^2 + y^2 = 100\).
Key Concepts
Circle EquationsCoordinate GeometryAlgebraic EquationsPrecalculus
Circle Equations
Understanding the standard form of the equation of a circle is crucial for solving problems in coordinate geometry. The general form is \(x-a)^2 + (y-b)^2 = r^2\), where \(a\) and \(b\) correspond to the x and y coordinates of the center of the circle, respectively, and \(r\) is the radius. By plugging in the given center and radius, we can derive the specific equation for any circle.
For example, for a circle with center at (-4,0) and a radius of 10, the equation becomes \(x+4)^2 + y^2 = 100\). Notice how the sign before the 4 is reversed in the equation; this is because subtracting a negative is the same as adding a positive. The right side of the equation, \(r^2\), turns our radius into \(10^2\) or 100.
For example, for a circle with center at (-4,0) and a radius of 10, the equation becomes \(x+4)^2 + y^2 = 100\). Notice how the sign before the 4 is reversed in the equation; this is because subtracting a negative is the same as adding a positive. The right side of the equation, \(r^2\), turns our radius into \(10^2\) or 100.
Coordinate Geometry
Coordinate geometry, also known as analytic geometry, is all about defining geometric figures using algebraic equations and a coordinate system. In the context of circle equations, coordinate geometry allows us to pinpoint the location of the circle's center and understand its size through the radius. The Cartesian coordinate system with its x and y axes is the most commonly used system for this purpose.
In solving our textbook problem, we start by identifying the center point of the circle at (-4,0), which lies on the x-axis, and represent the circle's size with the equation \(x+4)^2 + y^2 = 100\) based on its radius. This visual and mathematical bridging done by coordinate geometry makes it a powerful tool in many areas of math and science.
In solving our textbook problem, we start by identifying the center point of the circle at (-4,0), which lies on the x-axis, and represent the circle's size with the equation \(x+4)^2 + y^2 = 100\) based on its radius. This visual and mathematical bridging done by coordinate geometry makes it a powerful tool in many areas of math and science.
Algebraic Equations
Algebraic equations form the foundation of expressing mathematical ideas and solving problems. When dealing with circle equations in standard form, we apply algebra to manipulate and simplify expressions. Subtraction or addition of the same number on both sides, squaring binomials, and manipulating constants and variables are frequent operations we use to achieve the desired form.
To illustrate, when the textbook problem asked us to plug -4 and 0 into the standard equation of a circle, the signs and simplification we carried out were tools from algebra making it cleaner and easier to read. Resulting in \(x+4)^2 + y^2 = 100\), we used algebra to showcase the circle's properties in a neat and universally understood format.
To illustrate, when the textbook problem asked us to plug -4 and 0 into the standard equation of a circle, the signs and simplification we carried out were tools from algebra making it cleaner and easier to read. Resulting in \(x+4)^2 + y^2 = 100\), we used algebra to showcase the circle's properties in a neat and universally understood format.
Precalculus
Precalculus is a course that prepares students for the more challenging concepts they'll encounter in calculus. It covers a wide range of topics, including functions, complex numbers, and, of course, the geometry we're discussing here. The equation of a circle is a fundamental topic within precalculus because it combines elements of algebra and geometry, serving as a stepping stone to more advanced work in calculus.
Circle equations require a solid understanding of precalculus concepts to transition smoothly into the study of curves and limits in calculus. It's vital for students to be comfortable with these equations as they pave the way for modeling more complex systems and analyzing dynamic changes.
Circle equations require a solid understanding of precalculus concepts to transition smoothly into the study of curves and limits in calculus. It's vital for students to be comfortable with these equations as they pave the way for modeling more complex systems and analyzing dynamic changes.
Other exercises in this chapter
Problem 38
Evaluate each function at the given values of the independent variable and simplify. \(f(x)=\frac{|x+3|}{x+3}\) a. \(f(5)\) b. \(f(-5)\) c. \(f(-9-x)\)
View solution Problem 38
Use the given conditions to write an equation for each line in point-slope form and slope-intercept form. \(x\) -intercept \(=4\) and \(y\) -intercept \(=-2\)
View solution Problem 39
a. Why are the lines whose equations are \(y=\frac{1}{3} x+1\) and \(y=-3 x-2\) perpendicular? b. Use a graphing utility to graph the equations in a [-10,10,1]
View solution Problem 39
The table of values was generated by a graphing utility with a TABLE feature. Use the table to solve $$\begin{array}{|c|c|c|c|c|} \hline X & Y 1 & Y 2 & & \\ \h
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