Problem 34
Question
Graph each circle by hand if possible. Give the domain and range. $$(x+2)^{2}+(y-5)^{2}=16$$
Step-by-Step Solution
Verified Answer
Center: (-2, 5); Radius: 4; Domain: [-6, 2]; Range: [1, 9]
1Step 1: Identify Center and Radius
The equation of a circle is generally given by \((x-h)^2 + (y-k)^2 = r^2\), where \((h, k)\) is the center and \(r\) is the radius. For the given circle \((x + 2)^2 + (y - 5)^2 = 16\), we can rewrite it in terms of the standard equation: \((x - (-2))^2 + (y - 5)^2 = 4^2\). This reveals that the center \((h, k)\) is \((-2, 5)\) and the radius \(r\) is 4.
2Step 2: Graph the Circle
To graph the circle, start by plotting the center \((-2, 5)\) on the coordinate plane. From the center, measure a distance of 4 units in all directions (up, down, left, and right) to accommodate the radius. Mark these points, then draw a smooth curve connecting them to form the circle.
3Step 3: Determine the Domain
The domain of a circle involves the set of all possible \(x\)-values. In this case, since the leftmost point of the circle is at \((-2 - 4, 5)\) or \((-6, 5)\) and the rightmost point is at \((-2 + 4, 5)\) or \((2, 5)\), the domain is \([-6, 2]\).
4Step 4: Determine the Range
The range of a circle involves the set of all possible \(y\)-values. Here, the lowest point is at \((-2, 5 - 4)\) or \((-2, 1)\) and the highest point is at \((-2, 5 + 4)\) or \((-2, 9)\), making the range \([1, 9]\).
Key Concepts
Domain and RangeEquation of a CircleCenter and Radius
Domain and Range
The domain and range of a circle define all the possible values that the circle's equation can hold for the variables \(x\) and \(y\).
In simpler terms, the domain is the complete set of \(x\)-values (horizontal coordinates), while the range is the entire set of \(y\)-values (vertical coordinates).
Let's explore this using our given circle equation:
In simpler terms, the domain is the complete set of \(x\)-values (horizontal coordinates), while the range is the entire set of \(y\)-values (vertical coordinates).
Let's explore this using our given circle equation:
- Domain: Since the circle is centered at \((-2, 5)\) and has a radius of 4, it stretches horizontally from \(x = -2 - 4\) (leftmost point) to \(x = -2 + 4\) (rightmost point), giving us a domain of \([-6, 2]\).
- Range: Vertically, the circle extends from \(y = 5 - 4\) (lowest point) to \(y = 5 + 4\) (highest point), providing a range of \([1, 9]\).
Equation of a Circle
The equation of a circle is a fundamental concept in mathematics that allows us to represent a circle in a coordinate plane using a simple algebraic formula.
This standard equation is given by \[(x-h)^2 + (y-k)^2 = r^2\]where:
This standard equation is given by \[(x-h)^2 + (y-k)^2 = r^2\]where:
- \((h, k)\) represents the center of the circle.
- \(r\) represents the radius.
- Rewrite \((x+2)^2\) as \((x - (-2))^2\)\ and \((y-5)^2\){} remains unchanged.
- This shows the circle's center \((h, k) = (-2, 5)\)\ and the radius squared is 16, so \(r = 4\).
Center and Radius
The center and radius are two critical elements of a circle that define its unique position and scale on the coordinate plane.
Let's break down what each means and how we identify them in our equation.
The center of the circle is the point inside the circle that is equidistant from all points on the boundary.
Let's break down what each means and how we identify them in our equation.
The center of the circle is the point inside the circle that is equidistant from all points on the boundary.
- In the equation \((x-h)^2 + (y-k)^2 = r^2\), the values \(h\) and \(k\) pinpoint this center.
- For our example, \((x + 2)^2 + (y - 5)^2 = 16\), the center is located at \((-2, 5)\).
- In the same equation, \(r^2\) is the value you have on the right side, indicating the radius squared.
- So, for a radius squared of 16, the radius is \(r = 4\).
Other exercises in this chapter
Problem 34
Find an equation for each ellipse. Minor axis of length \(4 ;\) foci \((-5,0)\) and \((5,0)\)
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Determine the type of conic section represented by each equation, and graph it, provided a graph exists. $$-4 x^{2}+8 x+y^{2}+6 y=-6$$
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Find a rectangular equation. State the appropriate interval for \(x\) or \(y .\) $$x=\frac{1}{\sqrt{t+2}}, y=\frac{t}{t+2}, \text { for } t \text { in }(-2, \in
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Find an equation for each ellipse. Center \((5,2) ;\) minor axis vertical, with length \(8 ; c=3\)
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