Problem 35
Question
Write the standard form of the equation of the circle with the given center and radius. $$ \text { Center }(-1,4), r=2 $$
Step-by-Step Solution
Verified Answer
The standard form of the equation of the circle centered at (-1,4) with radius 2 is \[(x + 1)^2 + (y - 4)^2 = 4\].
1Step 1: Recognize the Standard Form of a Circle's Equation
The Standard Form of a Circle's Equation is \[ (x-a)^2 + (y-b)^2 = r^2 \], where the center of the circle is point (a,b) and r stands for the radius.
2Step 2: Apply the given values
We plug the given values inside the formula. The center is (-1,4), so a = -1 and b = 4. The radius r = 2. Substituting these into the equation, we get \[ (x - (-1))^2 + (y - 4)^2 = 2^2 \].
3Step 3: Simplify the equation
Now, simplify the equation as \[ (x + 1)^2 + (y - 4)^2 = 4 \]. This is the standard form of the equation of the given circle.
Key Concepts
Circle EquationsAlgebraic GeometryCoordinate Geometry
Circle Equations
Understanding the equation of a circle is a fundamental aspect of both algebraic and coordinate geometry. It represents the set of all points that maintain a constant distance, known as the radius, from a fixed point called the center.
The standard form of a circle's equation might look a little daunting at first, but it's simply giving us a blueprint for all the points that make up the circle. The standard form is written as \( (x-a)^2 + (y-b)^2 = r^2 \). Here, the point (a,b) represents the center of the circle on a coordinate plane, and 'r' is the radius.
When you're given the center and the radius of a circle, like in the example where the center is (-1,4) and the radius is 2, just substitute these values into the standard form. By adding 1 to 'x' and subtracting 4 from 'y', and then squaring the radius, we plot all the points around the center that are exactly 2 units away, forming our circle.
The standard form of a circle's equation might look a little daunting at first, but it's simply giving us a blueprint for all the points that make up the circle. The standard form is written as \( (x-a)^2 + (y-b)^2 = r^2 \). Here, the point (a,b) represents the center of the circle on a coordinate plane, and 'r' is the radius.
When you're given the center and the radius of a circle, like in the example where the center is (-1,4) and the radius is 2, just substitute these values into the standard form. By adding 1 to 'x' and subtracting 4 from 'y', and then squaring the radius, we plot all the points around the center that are exactly 2 units away, forming our circle.
Algebraic Geometry
Algebraic geometry is a branch of mathematics that blends techniques from abstract algebra, especially commutative algebra, with the language and the problems of geometry. It allows us to describe geometric forms like lines, curves, and shapes using algebraic expressions. This creates a powerful toolkit for solving a range of problems.
For circles, algebraic geometry lets us work out not just the position of points on the circumference but also other properties like tangents, sectors, and even the circle's area and perimeter. It's this blend of algebra and geometry that makes understanding concepts like the standard form of a circle's equation so enriching - it's where numbers meet shapes, marrying calculation with visualization.
For circles, algebraic geometry lets us work out not just the position of points on the circumference but also other properties like tangents, sectors, and even the circle's area and perimeter. It's this blend of algebra and geometry that makes understanding concepts like the standard form of a circle's equation so enriching - it's where numbers meet shapes, marrying calculation with visualization.
Coordinate Geometry
Coordinate geometry, also known as analytic geometry, is about describing geometric figures using a system of coordinates and the principles of algebra. This is the framework we use to plot points, lines, and curves on a graph.
The Cartesian coordinate system, with its x and y axes, allows us to take algebraic equations and give them a visual representation. Circles are perfect examples of how algebraic equations can create clear visual patterns on a graph. By using the standard form equation and identifying the values for the center and radius, one can effectively plot a circle on the coordinate plane.
This process not only helps visualize solutions but also provides insight into other related geometric properties, such as the relation of a point to the circle (inside, outside or on the circle) and finding the intersecting points with other geometric entities like lines or other circles.
The Cartesian coordinate system, with its x and y axes, allows us to take algebraic equations and give them a visual representation. Circles are perfect examples of how algebraic equations can create clear visual patterns on a graph. By using the standard form equation and identifying the values for the center and radius, one can effectively plot a circle on the coordinate plane.
This process not only helps visualize solutions but also provides insight into other related geometric properties, such as the relation of a point to the circle (inside, outside or on the circle) and finding the intersecting points with other geometric entities like lines or other circles.
Other exercises in this chapter
Problem 35
Find: a. \((f \circ g)(x)\) b. the domain of \(f \circ g\) $$f(x)=x^{2}+4, g(x)=\sqrt{1-x}$$
View solution Problem 35
In Exercises \(33-44\), find and simplify the difference quotient $$\frac{f(x+h)-f(x)}{h}, h \neq 0$$for the given function. $$f(x)=3 x+7$$
View solution Problem 36
Begin by graphing the standard cubic function, \(f(x)=x^{3} .\) Then use transformations of this graph to graph the given function. $$ g(x)=x^{3}-2 $$
View solution Problem 36
Use the given conditions to write an equation for each line in point-slope form and slope-intercept form. Passing through \((1,-3)\) with \(x\) -intercept \(=-1
View solution