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Chapter 3: Problem 25

Exer. 23-34: Sketch the graph of the circle or semicircle. $$ (x+3)^{2}+(y-2)^{2}=9 $$

Short Answer

Expert verified
Center: \((-3, 2)\), Radius: 3.

Step by step solution

01

Identify the Standard Equation Form

The given equation \((x+3)^2 + (y-2)^2 = 9\) is in the standard form of a circle \((x-h)^2 + (y-k)^2 = r^2\). Here, \((h, k)\) is the center of the circle and \(r\) is the radius.
02

Determine the Circle's Parameters

From comparing the given equation to the standard form, we identify the center of the circle \((h, k)\) as \((-3, 2)\) and the radius \(r\) as \( \sqrt{9} = 3\).
03

Sketch the Circle

First, plot the center of the circle at \((-3, 2)\). Then, using the radius of 3 units, draw the circle such that it is 3 units away in all directions from the center.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Graphing Circles
Graphing circles is straightforward once you understand the basics, rooted in the circle's equation. Imagine having a fixed center point. From this center, a circle is simply all the points at a fixed distance (known as the radius) away from this central location. In any coordinate system, you graph a circle by first plotting its center, then marking points that lie that fixed distance from it.

In the problem, the circle's center is (-3, 2), and with a radius of 3, you draw points that are 3 units away in every direction. To visualize:
  • Start at the center (-3, 2)
  • Move 3 units up, down, left, and right
  • Connect these points smoothly in a round shape
This simple process of graphing circles helps you understand how circles fit in algebra, and makes sketching easier each time!
Equations of Circles
The equation of a circle in a plane grid is a tidy way to describe all its parts. The standard equation \((x-h)^2 + (y-k)^2 = r^2\) gives us a circle's blueprint. Here:
  • \((h, k)\) is the center of the circle
  • \(r\) is the radius
Every part of this equation tells you something essential about the circle it represents. By comparing your circle鈥檚 equation to this standard form, you can identify the fundamental properties of the circle.

In our case, the equation \((x+3)^2 + (y-2)^2 = 9\) easily compares to the standard form. We can see that \(h = -3\), \(k = 2\), with \(r^2 = 9\). From this, deriving the radius and center becomes a simple task, making the equation highly useful for deeper understanding and precise graphing.
Radius and Center of a Circle
To explore a circle, you start with its center and radius. These two elements dictate everything about the circle's size and position. The center \((h, k)\) is like your circle's home base, while the radius is the defined distance from this home base to any point on the circle's curve.

In the specific equation \((x+3)^2 + (y-2)^2 = 9\), comparing it with the standard circle equation gives us:
  • Center \((-3, 2)\): Telling us where the circle sits on the grid
  • Radius \(r = 3\): Telling you how wide the circle stretches
A clear understanding of these two core aspects allows you to visualize and draw the circle swiftly and accurately. Recognizing how they connect to the equation empowers you to decode or devise any circle equation you encounter.

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Most popular questions from this chapter

Exer. 47-52: Sketch the graph of \(f\). $$ f(x)= \begin{cases}x-3 & \text { if } x \leq-2 \\ -x^{2} & \text { if }-2

Let \(y=f(x)\) be a function with domain \(D=[-2,6]\) and range \(R=[-4,8]\). Find the domain \(D\) and range \(R\) for each function. Assume \(f(2)=8\) and \(f(6)=-4\). (a) \(y=-2 f(x)\) (b) \(y=f\left(\frac{1}{2} x\right)\) (c) \(y=f(x-3)+1\) (d) \(y=f(x+2)-3\) (e) \(y=f(-x)\) (f) \(y=-f(x)\) (g) \(y=f(|x|)\) (h) \(y=|f(x)|\)

Exer. 27-32: If the point \(P\) is on the graph of a function \(f\), find the corresponding point on the graph of the given function. $$ P(3,-2) ; \quad y=2 f(x-4)+1 $$

Exer. 21-34: Find (a) \((f \circ g)(x)\) and the domain of \(f \circ g\) and (b) \((g \circ f)(x)\) and the domain of \(g \circ f\). $$ f(x)=\sqrt{3-x}, \quad g(x)=\sqrt{x^{2}-16} $$

Exer. 47-52: Sketch the graph of \(f\). $$ f(x)= \begin{cases}3 & \text { if } x \leq-1 \\ -2 & \text { if } x>-1\end{cases} $$
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