Problem 16
Question
A circle is described in words. Give its Cartesian equation. The circle with center at the origin and radius 2
Step-by-Step Solution
Verified Answer
The Cartesian equation is \\(x^2 + y^2 = 4\\).
1Step 1: Understand the General Equation of a Circle
The general equation of a circle with center \(h, k\) and radius \ r \ is given by: \((x - h)^2 + (y - k)^2 = r^2\). This equation represents all the points \(x, y\) that are equidistant \(r\) from the center \(h, k\).
2Step 2: Identify Given Values
In this problem, the circle's center is at the origin \(0, 0\) and the radius is 2. Thus, \(h = 0\), \(k = 0\), and \(r = 2\).
3Step 3: Substitute Values into General Equation
Substitute the values \(h = 0\), \(k = 0\), and \(r = 2\) into the general equation: \((x - 0)^2 + (y - 0)^2 = 2^2\). This simplifies to: \(x^2 + y^2 = 4\).
Key Concepts
General Equation of a CircleCenter of a CircleRadius of a Circle
General Equation of a Circle
In mathematics, the Cartesian equation of a circle is foundational for understanding how circles are represented in a plane. A circle in a Cartesian coordinate system can be described with the general equation:
It's important because it allows us to visualize and understand the position and distance properties of a circle relative to a coordinate plane.
By simply substituting the values for the center and radius into this equation, we can model any circle accurately. The beauty of this equation is its simplicity—once understood, it becomes a powerful tool for solving many geometrical problems related to circles.
- \((x - h)^2 + (y - k)^2 = r^2\)
It's important because it allows us to visualize and understand the position and distance properties of a circle relative to a coordinate plane.
By simply substituting the values for the center and radius into this equation, we can model any circle accurately. The beauty of this equation is its simplicity—once understood, it becomes a powerful tool for solving many geometrical problems related to circles.
Center of a Circle
The circle's center is a crucial part of its definition in the coordinate plane. In the general equation
In our example, the circle's center is at the origin, which means \(h = 0\) and \(k = 0\). This places the circle right in the center of the coordinate plane, which is a special and common location for the center of a circle in many problems to simplify calculations.
When the center is at the origin, the equation simplifies significantly, eliminating the \(h\) and \(k\) terms, thus reducing our equation to:
- \((x - h)^2 + (y - k)^2 = r^2\)
In our example, the circle's center is at the origin, which means \(h = 0\) and \(k = 0\). This places the circle right in the center of the coordinate plane, which is a special and common location for the center of a circle in many problems to simplify calculations.
When the center is at the origin, the equation simplifies significantly, eliminating the \(h\) and \(k\) terms, thus reducing our equation to:
- \(x^2 + y^2 = r^2\)
Radius of a Circle
The radius of a circle plays a fundamental role in determining the size and scale of the circle. In the general circle equation
In simple terms, imagine you draw a line from the center of the circle to any point on its edge—that line is your radius. In our specific example, the radius is 2, meaning the circle measures 2 units from the center in all directions.
To use this in the general equation, you substitute \(r\) with its value, squaring it for the equation. This results in the specific form:
- \((x - h)^2 + (y - k)^2 = r^2\)
In simple terms, imagine you draw a line from the center of the circle to any point on its edge—that line is your radius. In our specific example, the radius is 2, meaning the circle measures 2 units from the center in all directions.
To use this in the general equation, you substitute \(r\) with its value, squaring it for the equation. This results in the specific form:
- \(x^2 + y^2 = 4\)
Other exercises in this chapter
Problem 16
Write the slope-intercept equation of the line determined by the given data. Slope \(0, y\) -intercept 3
View solution Problem 16
Plot several points, and sketch the graph of the function defined by the given expression. $$ \sqrt{x^{2}} $$
View solution Problem 16
Write the set using interval notation. Use the symbol \(\cup\) where appropriate. \(\\{u:|u-4| \geq 6\\}\)
View solution Problem 17
Find a function \(g\) such that \(h=g \circ f\) \(h(x)=\left(x^{2}+1\right) /\left(x^{4}+2 x^{2}+3\right), f(x)=x^{2}+1\)
View solution