Problem 29
Question
Write the equation of a circle in standard form with the following properties. Center at \((0,0) ;\) radius \(\frac{1}{4}\)
Step-by-Step Solution
Verified Answer
The equation of the circle is \(x^2 + y^2 = \frac{1}{16}\).
1Step 1: Understanding the Standard Form of a Circle
The standard form of an equation of a circle with center \((h, k)\) and radius \(r\) is \((x - h)^2 + (y - k)^2 = r^2\). In this equation, \((h, k)\) are the coordinates of the circle's center, and \(r\) is the radius.
2Step 2: Identify Circle's Center and Radius
From the problem, we know the center of the circle is \((0,0)\) and the radius is \(\frac{1}{4}\). Thus, \(h = 0\), \(k = 0\), and \(r = \frac{1}{4}\).
3Step 3: Substitute Values into Standard Form
Substitute \(h = 0\), \(k = 0\), and \(r = \frac{1}{4}\) into the standard form: \((x - 0)^2 + (y - 0)^2 = (\frac{1}{4})^2\). This simplifies to \(x^2 + y^2 = \frac{1}{16}\).
4Step 4: Simplify the Equation
The simplified equation of the circle is \(x^2 + y^2 = \frac{1}{16}\). This represents a circle centered at the origin with a radius of \(\frac{1}{4}\).
Key Concepts
Understanding the Standard Form of a CircleDetermining the Center of a CircleDetermining the Radius of a Circle
Understanding the Standard Form of a Circle
To understand the geometry of a circle, we start with its equation in standard form. The standard form of an equation of a circle helps to identify key features easily, such as the center and radius. This form is expressed as \[(x - h)^2 + (y - k)^2 = r^2\]Here is what each component means:
- \( (h, k) \) are the coordinates of the center of the circle. These two numbers tell us exactly where the center is on the coordinate plane.
- \( r \) represents the radius of the circle. This number indicates how wide the circle is.
Determining the Center of a Circle
The center of a circle is a vital piece of information that tells us where the circle is situated on a coordinate plane. In the context of the standard form of a circle's equation \((x-h)^2 + (y-k)^2 = r^2\), the coordinates \( (h, k) \) represent the center.
To find the center from an equation already in standard form, simply identify these coordinates. For example, in the equation given by \( (x - 0)^2 + (y - 0)^2 = r^2 \), it is clear that the center is at \((0, 0)\), also known as the origin.
Visualizing this, imagine plotting the point \( (0, 0) \) on a graph. This center remains unaffected by the value of the radius, which only defines the distance from this central point to any point on the edge of the circle.
Understanding the center not only helps with graphing the circle but is also crucial when solving systems involving multiple circles, as it provides insight into their placement relative to each other.
To find the center from an equation already in standard form, simply identify these coordinates. For example, in the equation given by \( (x - 0)^2 + (y - 0)^2 = r^2 \), it is clear that the center is at \((0, 0)\), also known as the origin.
Visualizing this, imagine plotting the point \( (0, 0) \) on a graph. This center remains unaffected by the value of the radius, which only defines the distance from this central point to any point on the edge of the circle.
Understanding the center not only helps with graphing the circle but is also crucial when solving systems involving multiple circles, as it provides insight into their placement relative to each other.
Determining the Radius of a Circle
The radius of a circle is the distance from its center to any point on its boundary, and it is a defining feature in the standard form equation. In the standard form \((x - h)^2 + (y - k)^2 = r^2\), the radius is represented by \( r \). This number is always positive and gives us an immediate sense of the circle's size.
Finding the radius from the equation involves taking the square root of the right side of the equation. If the equation is \( (x-h)^2 + (y-k)^2 = (\frac{1}{4})^2 \), then the radius \( r \) is the square root of \( \frac{1}{16} \), which simplifies to \( \frac{1}{4} \). Remember, the radius gives us the length from the center to the edge, which helps in sketching the circle or solving problems relating to its area and circumference.
By understanding how to derive the radius, you can unlock further equations involving the circle, like those for the circle's circumference \( (2\pi r) \)or area \( (\pi r^2) \),improving your ability to work with circles in all sorts of geometric contexts.
Finding the radius from the equation involves taking the square root of the right side of the equation. If the equation is \( (x-h)^2 + (y-k)^2 = (\frac{1}{4})^2 \), then the radius \( r \) is the square root of \( \frac{1}{16} \), which simplifies to \( \frac{1}{4} \). Remember, the radius gives us the length from the center to the edge, which helps in sketching the circle or solving problems relating to its area and circumference.
By understanding how to derive the radius, you can unlock further equations involving the circle, like those for the circle's circumference \( (2\pi r) \)or area \( (\pi r^2) \),improving your ability to work with circles in all sorts of geometric contexts.
Other exercises in this chapter
Problem 28
Graph each equation. \(\frac{(x-6)^{2}}{36}+\frac{(y+6)^{2}}{144}=1\)
View solution Problem 29
Solve each system of equations by substitution for real values of \(x\) and \(y.\) See Examples 2 and 3. $$ \left\\{\begin{array}{l} x^{2}+y^{2}=30 \\ y=x^{2} \
View solution Problem 29
Graph each equation. \((x+1)^{2}+4(y+2)^{2}=4\)
View solution Problem 30
Solve each system of equations by substitution for real values of \(x\) and \(y.\) See Examples 2 and 3. $$ \left\\{\begin{array}{l} x^{2}+y^{2}=20 \\ y=x^{2} \
View solution