Problem 26
Question
Find a formula that states that \(P(x, y)\) is a distance \(r>0\) from a fixed point \(C(h, k)\). Describe the set of all such points.
Step-by-Step Solution
Verified Answer
The formula is \((x - h)^2 + (y - k)^2 = r^2\), representing a circle with center \(C(h, k)\) and radius \(r\).
1Step 1: Understand the Problem
We are tasked with finding a formula that represents all points \(P(x, y)\) that are at a distance \(r\) from a fixed point \(C(h, k)\). This is the definition of a circle, where \((h, k)\) is the center and \(r\) is the radius.
2Step 2: Use the Distance Formula
To find the distance between two points \((x_1, y_1)\) and \((x_2, y_2)\), we use the distance formula: \(d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\).
3Step 3: Apply Distance to Circle Context
Since we want to find the set of points \(P(x, y)\) that are distance \(r\) from \(C(h, k)\), we set up the equation using the distance formula: \(\sqrt{(x - h)^2 + (y - k)^2} = r\).
4Step 4: Square Both Sides
To eliminate the square root, square both sides of the equation: \((x - h)^2 + (y - k)^2 = r^2\).
5Step 5: Identify the Formula
The equation \((x - h)^2 + (y - k)^2 = r^2\) is the formula for a circle centered at \((h, k)\) with radius \(r\).
6Step 6: Describe the Set of Points
The set of all points \((x, y)\) that satisfy \((x - h)^2 + (y - k)^2 = r^2\) forms a circle with center \(C(h, k)\) and radius \(r\).
Key Concepts
Distance FormulaCenter of a CircleRadius of a Circle
Distance Formula
The distance formula is a pivotal mathematical tool that helps us determine the distance between two points in a plane. If you have two points, \(P_1(x_1, y_1)\) and \(P_2(x_2, y_2)\), the distance \(d\) between them is calculated using the formula:
This is crucial because it lays the groundwork for understanding the equation of a circle in a coordinate plane.
In the context of a circle, our two points are:
- \[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]
This is crucial because it lays the groundwork for understanding the equation of a circle in a coordinate plane.
In the context of a circle, our two points are:
- a fixed point \(C(h, k)\) (the center of the circle),
- and a point on the circle \(P(x, y)\).
Center of a Circle
The center of a circle is a critical concept when studying circles. In a coordinate plane, it is usually denoted by \(C(h, k)\).
The center controls where the circle is located. Think of it as the circle's anchor point.
Understanding the center is essential because any changes to \(h\) or \(k\) will shift the entire circle horizontally or vertically across the plane.
The center controls where the circle is located. Think of it as the circle's anchor point.
Understanding the center is essential because any changes to \(h\) or \(k\) will shift the entire circle horizontally or vertically across the plane.
- Positioned at \(C(h, k)\), it serves as the reference point from which every point on the circle is equidistant.
- This distance is known as the radius.
Radius of a Circle
The radius of a circle, usually represented as \(r\), is the distance from the center to any point on the circle. In a circle's equation, this distance is a constant and is fundamental to defining the circle's size.
When you write the equation \((x - h)^2 + (y - k)^2 = r^2\), the radius is the \(r\) part:
Every point \(P(x, y)\) on the circle maintains this constant distance \(r\) from the center, creating the circular shape we are familiar with.
When you write the equation \((x - h)^2 + (y - k)^2 = r^2\), the radius is the \(r\) part:
- \(r\) tells you how far out the circle extends from the center point \(C(h, k)\).
- The larger the \(r\), the bigger the circle.
Every point \(P(x, y)\) on the circle maintains this constant distance \(r\) from the center, creating the circular shape we are familiar with.
Other exercises in this chapter
Problem 26
Exer. 21-32: Find the domain of \(f\). $$ f(x)=\frac{4 x}{6 x^{2}+13 x-5} $$
View solution Problem 26
Exer. 23-34: Sketch the graph of the circle or semicircle. $$ (x-4)^{2}+(y+2)^{2}=4 $$
View solution Problem 27
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
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Exer. 21-32: Find a general form of an equation of the line through the point \(A\) that satisfies the given condition. $$ A(4,-5) ; \quad \text { through } B(-
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