Elementary Geometry For College Students, 7e
7th Edition
ISBN:9781337614085
Author:Alexander, Daniel C.; Koeberlein, Geralyn M.
Publisher:Alexander, Daniel C.; Koeberlein, Geralyn M.
ChapterP: Preliminary Concepts
SectionP.CT: Test
Problem 1CT
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Question
Identify each part of the circle given its equation
![### Circle Equation and Properties
Given the standard form of a circle's equation:
\[ x^2 + (y + 10)^2 = 42.25 \]
#### Center
The center of the circle is at:
\[ (x, y) = (0, -10) \]
#### Diameter
To determine the diameter, we first need to find the radius \( r \). The given equation is in the form \( (x - h)^2 + (y - k)^2 = r^2 \), where \( (h, k) \) is the center and \( r \) is the radius. Here, \( r^2 = 42.25 \), so:
\[ r = \sqrt{42.25} = 6.5 \]
Therefore, the diameter \( D \) is twice the radius:
\[ D = 2 \times 6.5 = 13 \]
So, the diameter of the circle is:
\[ \text{Diameter}: 13 \]
These calculations help us understand the key properties of the circle defined by the given equation.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fa7596019-6a00-40b6-acbc-de98b6fa36ba%2F11c076ae-04bc-4697-b738-380fbc36c184%2F4abm4ub_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Circle Equation and Properties
Given the standard form of a circle's equation:
\[ x^2 + (y + 10)^2 = 42.25 \]
#### Center
The center of the circle is at:
\[ (x, y) = (0, -10) \]
#### Diameter
To determine the diameter, we first need to find the radius \( r \). The given equation is in the form \( (x - h)^2 + (y - k)^2 = r^2 \), where \( (h, k) \) is the center and \( r \) is the radius. Here, \( r^2 = 42.25 \), so:
\[ r = \sqrt{42.25} = 6.5 \]
Therefore, the diameter \( D \) is twice the radius:
\[ D = 2 \times 6.5 = 13 \]
So, the diameter of the circle is:
\[ \text{Diameter}: 13 \]
These calculations help us understand the key properties of the circle defined by the given equation.
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