Problem 25
Question
Exer. 23-34: Sketch the graph of the circle or semicircle. $$ (x+3)^{2}+(y-2)^{2}=9 $$
Step-by-Step Solution
Verified Answer
Center: \((-3, 2)\), Radius: 3.
1Step 1: 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.
2Step 2: 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 \( \sqrt{9} = 3\).
3Step 3: 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.
Key Concepts
Graphing CirclesEquations of CirclesRadius and Center of a Circle
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:
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
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:
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.
- \((h, k)\) is the center of the circle
- \(r\) is the radius
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:
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
Other exercises in this chapter
Problem 25
Exer. 21-32: Find a general form of an equation of the line through the point \(A\) that satisfies the given condition. $$ A(4,0) ; \quad \text { slope }-3 $$
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Exer. 21-32: Find the domain of \(f\). $$ f(x)=\frac{x+1}{x^{3}-4 x} $$
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Find a formula that expresses the fact that \(P(x, y)\) is a distance 5 from the origin. Describe the set of all such points.
View solution Problem 26
Exer. 13-26: Sketch, on the same coordinate plane, the graphs of \(f\) for the given values of \(c\). (Make use of symmetry, shifting, stretching, compressing,
View solution