Table of contents
- 0. Review of Algebra4h 16m
- 1. Equations & Inequalities3h 18m
- 2. Graphs of Equations43m
- 3. Functions2h 17m
- 4. Polynomial Functions1h 44m
- 5. Rational Functions1h 23m
- 6. Exponential & Logarithmic Functions2h 28m
- 7. Systems of Equations & Matrices4h 6m
- 8. Conic Sections2h 23m
- 9. Sequences, Series, & Induction1h 19m
- 10. Combinatorics & Probability1h 45m
8. Conic Sections
Circles
Problem 9
Textbook Question
Without actually graphing, identify the type of graph that each equation has.
x2+y2=144
![](/channels/images/assetPage/verifiedSolution.png)
1
Recognize the standard form of the equation: The given equation is \(x^2 + y^2 = 144\). This is a form of the equation \(x^2 + y^2 = r^2\), which is the standard form of a circle centered at the origin.
Identify the components of the equation: In the equation \(x^2 + y^2 = 144\), the terms \(x^2\) and \(y^2\) indicate that both variables are squared and have the same coefficient, which is 1 in this case.
Determine the radius of the circle: The equation \(x^2 + y^2 = r^2\) represents a circle with radius \(r\). Here, \(r^2 = 144\), so the radius \(r\) is the square root of 144.
Calculate the radius: The square root of 144 is 12, so the radius of the circle is 12.
Conclude the type of graph: Since the equation is in the form \(x^2 + y^2 = r^2\) and represents a circle with a radius of 12, the graph of this equation is a circle centered at the origin with a radius of 12.
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