Problem 44
Question
Give the center and radius of the circle described by the equation and graph each equation. $$ (x-2)^{2}+(y-3)^{2}=16 $$
Step-by-Step Solution
Verified Answer
The center of the circle is (2,3) and the radius is 4.
1Step 1: Identify the circle's center
The center (h,k) can be identified by looking at the terms in the parentheses in the given equation. In the equation \((x-2)^2 + (y-3)^2 = 16\), the center (h,k) is (2,3).
2Step 2: Identify the circle's radius
The radius can be identified by taking the square root of the number on the right-hand side of the equation. In this case, the square root of 16 is 4. So, the radius of the circle is 4.
3Step 3: Graph the circle
To graph the circle, locate the center point (2,3) on an x-y coordinate grid. Then, draw a circle with a radius of 4 units centered at that point. Be sure to use a consistent scale so that the circle appears round rather than elliptical.
Key Concepts
Understanding the Circle's CenterCalculating the Circle's RadiusGraphing the Circle
Understanding the Circle's Center
The center of a circle is a crucial point that helps define its location on a coordinate plane. To find it for a given circle's equation, observe the structure of the equation, specifically the expression within the parentheses.
For example, consider the equation \[(x-2)^2 + (y-3)^2 = 16.\]The center of this circle, denoted as \((h, k)\), is found by analyzing the values subtracted from each variable within the parentheses. Here, \(h = 2\) and \(k = 3\), making the center \((2, 3)\).
For example, consider the equation \[(x-2)^2 + (y-3)^2 = 16.\]The center of this circle, denoted as \((h, k)\), is found by analyzing the values subtracted from each variable within the parentheses. Here, \(h = 2\) and \(k = 3\), making the center \((2, 3)\).
- The expression \((x - h)^2 + (y - k)^2\) helps locate the circle's center \((h, k)\).
- The signs inside the parentheses are reversed. So, if the equation is \((x - 2)^2\), the \(x\)-coordinate of the center is 2.
- Always match the values from inside the parentheses to obtain the circle's center.
Calculating the Circle's Radius
The radius of a circle represents the distance from its center to any point on its circumference. In mathematical terms, it is half the diameter of the circle.
When given an equation of a circle, like \((x-2)^2 + (y-3)^2 = 16\), identifying the radius involves focusing on the number on the equation's right-hand side. This number represents the radius squared.
When given an equation of a circle, like \((x-2)^2 + (y-3)^2 = 16\), identifying the radius involves focusing on the number on the equation's right-hand side. This number represents the radius squared.
- To find the radius, you take the square root of this number.
- In our example, the right side is 16. Thus, \(\sqrt{16} = 4\).
- This calculation shows that the circle's radius is 4 units.
Graphing the Circle
Successfully graphing a circle involves a few basic steps, which begin with understanding the circle's equation. Once you have determined the circle's center and radius, you can accurately plot it on a coordinate grid.
- Start by marking the center point on the graph. For our example, the center \((2, 3)\) is plotted by moving 2 units along the x-axis and 3 units along the y-axis.
- From this center, measure out the radius, which is 4 units in this case, in all directions to establish the circumference.
- Using a compass or freehand, draw a circle around the center point that touches each radius endpoint.
Other exercises in this chapter
Problem 44
Express the given function h as a composition of two functions f and g so that \(h(x)=(f \circ g)(x)\) $$h(x)=|3 x-4|$$
View solution Problem 44
In Exercises \(33-44\), find and simplify the difference quotient $$\frac{f(x+h)-f(x)}{h}, h \neq 0$$for the given function. $$f(x)=\frac{1}{2 x}$$
View solution Problem 45
Give the slope and y-intercept of each line whose equation is given. Then graph the line. $$y=-\frac{3}{5} x+7$$
View solution Problem 45
Evaluate each piecewise function at the given values of the independent variable. $$\begin{aligned} &f(x)=\left\\{\begin{array}{ll} 3 x+5 & \text { if } x
View solution