Problem 27
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 \((x - h)^2 + (y - k)^2 = r^2\) describes a circle with center \(C(h, k)\) and radius \(r\).
1Step 1: Understand the Problem
We need to find a formula that describes all points \((x, y)\) that are at a distance \(r\) from a fixed point \(C(h, k)\). This set of points forms a geometric figure.
2Step 2: Use the Distance Formula
The distance between two points \((x_1, y_1)\) and \((x_2, y_2)\) is given by the formula \(d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\). Here, \((x_1, y_1)\) is \((h, k)\) and \((x_2, y_2)\) is \((x, y)\).
3Step 3: Set Up the Equation
According to the problem, the distance \(d\) is \(r\). Thus, we set up the equation: \[\sqrt{(x - h)^2 + (y - k)^2} = r\]
4Step 4: Eliminate the Square Root
Square both sides of the equation to remove the square root, yielding: \((x - h)^2 + (y - k)^2 = r^2\).
5Step 5: Interpret the Result
The equation \((x - h)^2 + (y - k)^2 = r^2\) represents a circle centered at point \(C(h, k)\) with radius \(r\). All points \((x, y)\) that satisfy this equation are on the circumference of the circle.
Key Concepts
Distance FormulaCoordinate GeometryEquation of a Circle
Distance Formula
Understanding the distance formula is crucial in coordinate geometry as it helps determine the distance between two points on a coordinate plane. To find this, we use the formula:
- \(d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\)
- \((x_1, y_1) = (h, k)\), the fixed center of the circle
- \((x_2, y_2) = (x, y)\), any point on the circle's circumference
Coordinate Geometry
Coordinate geometry, also known as analytic geometry, provides a powerful toolset for exploring geometric concepts through algebra. This branch of mathematics uses coordinates on the Cartesian plane to represent geometrical figures, like lines, curves, and here, circles.
By transforming geometric shapes into equations, coordinate geometry allows us to manipulate and analyze these shapes algebraically.
In our context, a circle is a set of all points equidistant from a center point \((h, k)\).
Representing a circle algebraically allows for:
By transforming geometric shapes into equations, coordinate geometry allows us to manipulate and analyze these shapes algebraically.
In our context, a circle is a set of all points equidistant from a center point \((h, k)\).
Representing a circle algebraically allows for:
- Easy determination of whether a point lies on the circle.
- The ability to find the intersection points with other geometric figures.
- Calculation of properties like diameter and area.
Equation of a Circle
The equation \((x - h)^2 + (y - k)^2 = r^2\) is a standard form for representing circles. Understanding each component of this equation:
- \((h, k)\) represents the circle's center
- \(r\) is the radius, which defines the distance from the center to the circle's edge
- Use the distance formula to equate distance, \(r\), between the center \((h, k)\) and any point \((x, y)\)
- Square both sides of the equation to remove the square root, simplifying computation
Other exercises in this chapter
Problem 27
Exer. 21-32: Find the domain of \(f\). $$ f(x)=\frac{\sqrt{2 x-3}}{x^{2}-5 x+4} $$
View solution Problem 27
Exer. 23-34: Sketch the graph of the circle or semicircle. $$ (x+3)^{2}+y^{2}=16 $$
View solution Problem 28
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,-1) ; \quad y=2 f(x
View solution Problem 28
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)=x^{3}+5, \quad g(x)
View solution