Chapter 12: Problem 20
Find the center and radius for each circle. $$5 x^{2}+5 y^{2}=5$$
Short Answer
Expert verified
Center: (0,0), Radius: 1
Step by step solution
01
Understand the given equation
The given equation of the circle is \[5x^{2} + 5y^{2} = 5\].
02
Divide the entire equation by 5
To simplify the equation, divide every term by 5 to get the equation in standard form: \[x^{2} + y^{2} = 1\].
03
Identify the standard form of a circle
This is now in the standard form of a circle equation: \[(x - h)^{2} + (y - k)^{2} = r^{2}\], where (h, k) is the center and r is the radius.
04
Determine the center
From the equation \[x^{2} + y^{2} = 1\], it is clear that (h, k) = (0, 0). The center of the circle is at the origin.
05
Find the radius
Compare the equation \[x^{2} + y^{2} = 1\] with the standard form \[(x - h)^{2} + (y - k)^{2} = r^{2}\]. You see that \(r^{2} = 1\), and therefore the radius \(r = \pm 1\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Standard Form of a Circle
Understanding the standard form of a circle's equation is crucial in solving problems related to circles. The equation of a circle in standard form is: \ \[ (x - h)^{2} + (y - k)^{2} = r^{2} \] Here, \( (h, k) \) represents the coordinates of the center of the circle, and \( r \) is the radius. This form helps identify the circle's key features, such as its center and radius, quickly and easily.
When given a circle's equation not in this form, like \( 5x^2 + 5y^2 = 5 \), you need to convert it to the standard form first. This involves simplifying the equation appropriately. For instance, by dividing each term by 5, the equation becomes \( x^2 + y^2 = 1 \), which now fits the standard form better and makes it simpler to identify the circle's center and radius.
When given a circle's equation not in this form, like \( 5x^2 + 5y^2 = 5 \), you need to convert it to the standard form first. This involves simplifying the equation appropriately. For instance, by dividing each term by 5, the equation becomes \( x^2 + y^2 = 1 \), which now fits the standard form better and makes it simpler to identify the circle's center and radius.
Center of a Circle
The center of a circle in its standard equation form \( (x - h)^{2} + (y - k)^{2} = r^{2} \) is given by the coordinates \( (h, k) \). In this normalized equation, the expressions \( x - h \) and \( y - k \) indicate horizontal and vertical shifts from the origin respectively.
For example, in the equation \( x^2 + y^2 = 1 \), which is derived from the original problem, the values \( h \) and \( k \) are both 0. Therefore, the center of this circle is at the coordinate \( (0, 0) \), meaning it is located at the origin.
For example, in the equation \( x^2 + y^2 = 1 \), which is derived from the original problem, the values \( h \) and \( k \) are both 0. Therefore, the center of this circle is at the coordinate \( (0, 0) \), meaning it is located at the origin.
- Essentially, any circle equation where neither \( x \) nor \( y \) is shifted (i.e., no \( h \) or \( k \)) has its center at the origin.
- If the terms involve \( (x - 3) \) or \( (y + 2) \), for instance, the center would then be at \( (3, -2) \).
Radius of a Circle
The radius is a fundamental aspect of a circle. In the standard form \( (x - h)^{2} + (y - k)^{2} = r^{2} \), the term \( r^2 \) represents the square of the radius. To find the actual radius, you take the square root of this value.
In our example equation \( x^2 + y^2 = 1 \), the term on the right side is 1. Hence, we have \( r^2 = 1 \). Taking the square root, \( r \) is \( \ pm 1 \). This means the radius of the circle is 1 unit.
In our example equation \( x^2 + y^2 = 1 \), the term on the right side is 1. Hence, we have \( r^2 = 1 \). Taking the square root, \( r \) is \( \ pm 1 \). This means the radius of the circle is 1 unit.
- Radius should always be a positive value as it is a distance measurement.
- The radius tells you the distance from any point on the circle to its center.
- When interpreting the standard form, always remember to take the square root of the right-hand term to get the radius.