Problem 34
Question
write the standard form of the equation of the circle with the given center and radius. $$ \text { Center }(2,-1), r=4 $$
Step-by-Step Solution
Verified Answer
The standard form of the equation of the circle with given center (2,-1) and radius 4 is \( (x-2)^2 + (y+1)^2 = 16 \).
1Step 1: Identify the given values
From the given, we can see that the center of the circle is at (h,k) = (2,-1) and the radius r = 4.
2Step 2: Substitute the given values into the standard form
Substitute h = 2, k = -1, and r = 4 into the standard form equation of a circle \( (x-h)^2 + (y-k)^2 = r^2 \). This gives us \( (x-2)^2 + (y-(-1))^2 = 4^2 \).
3Step 3: Simplify the equation
Simplify the equation to get \( (x-2)^2 + (y+1)^2 = 16 \).
Key Concepts
Equation of a CircleCircle Center and RadiusAlgebraic Equations
Equation of a Circle
Understanding the equation of a circle is crucial in geometry. Imagine a circle as a series of points that are all at the same distance from a central point. This distance is the radius, and the central point is the center of the circle. Mathematically, the standard form equation that represents a circle in a coordinate plane is written as:
\[ (x - h)^2 + (y - k)^2 = r^2 \]
In this equation, \( (h, k) \) represents the coordinates of the center of the circle, and \( r \) denotes the radius. The variables \( x \) and \( y \) correspond to the coordinates of any point on the circle. Substituting the specific values from an exercise, the equation creates a representation that defines the entire circle.
\[ (x - h)^2 + (y - k)^2 = r^2 \]
In this equation, \( (h, k) \) represents the coordinates of the center of the circle, and \( r \) denotes the radius. The variables \( x \) and \( y \) correspond to the coordinates of any point on the circle. Substituting the specific values from an exercise, the equation creates a representation that defines the entire circle.
Circle Center and Radius
For a circle in algebra, the key components include its center and radius. The center, denoted as \( (h, k) \), is the fixed point from which every point on the circle's perimeter is equidistant. This spot acts like an anchor for the circle's shape.
Understanding the Radius
The radius, usually just a positive number \( r \), is that constant distance between the center and any point on the circle's boundary. It's the defining measure that controls the size of the circle – the larger the radius, the larger the circle. When you're given these two pieces of information, you have all you need to delineate a circle's shape and position on a coordinate plane. Identifying the correct center and radius from a problem, then correctly placing them into the equation, is a fundamental skill in geometry.Algebraic Equations
Algebraic equations are the bread and butter of algebra. They are statements of equality containing algebraic expressions that include variables and constants. These equations allow us to solve for unknowns and model real-life scenarios mathematically.
Equations of circles are a specific type of algebraic equation. In the case of the circle, the unknowns are the coordinates \( x \) and \( y \) of points that lie on the circumference, and the constants are the coordinates of the center and the radius of the circle. Through the process of simplifying these equations, we can express a wealth of geometric relations algebraically. As with any equation, balance is key - what you do to one side must also be done to the other to maintain equality. This foundational concept facilitates every step in solving and manipulating algebraic equations, including those that define the properties of shapes like circles.
Equations of circles are a specific type of algebraic equation. In the case of the circle, the unknowns are the coordinates \( x \) and \( y \) of points that lie on the circumference, and the constants are the coordinates of the center and the radius of the circle. Through the process of simplifying these equations, we can express a wealth of geometric relations algebraically. As with any equation, balance is key - what you do to one side must also be done to the other to maintain equality. This foundational concept facilitates every step in solving and manipulating algebraic equations, including those that define the properties of shapes like circles.
Other exercises in this chapter
Problem 33
If you know a point on a line and you know the equation of a line perpendicular to this line, explain how to write the line's equation.
View solution Problem 33
Use the given conditions to write an equation for each line in point-slope form and slope-intercept form. Passing through \((-3,-1)\) and \((4,-1)\)
View solution Problem 34
Evaluate each function at the given values of the independent variable and simplify. $$ f(r)=\sqrt{25-r}-6 $$ a. \(f(16)\) b. \(f(-24) \) c. \(f(25-2 x)\)
View solution Problem 34
Find \(f+g, f-g,\) fg, and \(\frac{f}{g} .\) Determine the domain for each function. $$f(x)=x-6, g(x)=5 x^{2}$$
View solution