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
Related questions
Question
![On this page, we explore the standard form of a circle equation. The given equation is
\[3x^2 + 3y^2 - 24x - 6y + 3 = 0\]
This equation represents a circle with a center at \((h, k)\) and a radius \(r\). To find the values of \(h\), \(k\), and \(r\), we will rewrite this equation in the standard form.
**Rewriting the Equation**
The general equation of a circle in standard form is
\[(x - h)^2 + (y - k)^2 = r^2.\]
Initially, we can rewrite the given equation by grouping the \(x\) and \(y\) terms together:
\[3(x^2 - 8x) + 3(y^2 - 2y) = -3.\]
We will complete the square for the \(x\) and \(y\) terms. For that, let's factor out the 3 from the equation:
\[3[(x^2 - 8x) + (y^2 - 2y)] = -3.\]
Next, we complete the square:
1. Complete the square for \(x\):
\[x^2 - 8x + 16 = (x - 4)^2 - 16 + 16.\]
2. Complete the square for \(y\):
\[y^2 - 2y + 1 = (y - 1)^2 - 1 + 1.\]
Thus,
\[3[(x - 4)^2 - 16 + (y - 1)^2 - 1] = -3.\]
Simplify and balance the equation:
\[3[(x - 4)^2 - 16 + (y - 1)^2 - 1] = -3\]
\[3(x - 4)^2 - 48 + 3(y - 1)^2 - 3 = -3\]
\[3(x - 4)^2 + 3(y - 1)^2 - 51 = -3\]
\[3(x - 4)^2 + 3(y - 1)^2 - 51 = -3\]
\[3[(x - 4)^2 + (y - 1)^2] = 48.\]
\[(x - 4)^](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F04ddd72f-30f0-437a-bf8b-c24526128137%2Fea4c9700-6c14-462b-9c04-3c7b2e1ca52a%2F19n04ss_processed.png&w=3840&q=75)
Transcribed Image Text:On this page, we explore the standard form of a circle equation. The given equation is
\[3x^2 + 3y^2 - 24x - 6y + 3 = 0\]
This equation represents a circle with a center at \((h, k)\) and a radius \(r\). To find the values of \(h\), \(k\), and \(r\), we will rewrite this equation in the standard form.
**Rewriting the Equation**
The general equation of a circle in standard form is
\[(x - h)^2 + (y - k)^2 = r^2.\]
Initially, we can rewrite the given equation by grouping the \(x\) and \(y\) terms together:
\[3(x^2 - 8x) + 3(y^2 - 2y) = -3.\]
We will complete the square for the \(x\) and \(y\) terms. For that, let's factor out the 3 from the equation:
\[3[(x^2 - 8x) + (y^2 - 2y)] = -3.\]
Next, we complete the square:
1. Complete the square for \(x\):
\[x^2 - 8x + 16 = (x - 4)^2 - 16 + 16.\]
2. Complete the square for \(y\):
\[y^2 - 2y + 1 = (y - 1)^2 - 1 + 1.\]
Thus,
\[3[(x - 4)^2 - 16 + (y - 1)^2 - 1] = -3.\]
Simplify and balance the equation:
\[3[(x - 4)^2 - 16 + (y - 1)^2 - 1] = -3\]
\[3(x - 4)^2 - 48 + 3(y - 1)^2 - 3 = -3\]
\[3(x - 4)^2 + 3(y - 1)^2 - 51 = -3\]
\[3(x - 4)^2 + 3(y - 1)^2 - 51 = -3\]
\[3[(x - 4)^2 + (y - 1)^2] = 48.\]
\[(x - 4)^
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