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Core Connections Geometry, 2013
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Core Connections Geometry, 2013 View details
1. Section 12.1
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Exercise 6 Page 733

Practice makes perfect
a Consider the general equation of a circle.

(x- a)^2+(y- b)^2= r^2 [-1em] Midpoint:& ( a, b) Radius:& rFrom our diagram we can identify the midpoint and radius.

Let's substitute this information in the general formula. (x- 0)^2+(y- 0)^2= 3^2 [-1em] Midpoint:& ( 0, 0) Radius:& 3 This simplifies to x^2+y^2=9.
b If we compare the given equation to the general equation of a circle, we notice that it has a midpoint of (0,0) and a radius of 7, since 7^2=49.
(x- 0)^2+(y- 0)^2= 7^2 [-1em] Midpoint:& ( 0, 0) Radius:& 7 Now we can sketch the graph.

Subchapter links expand_more
arrow_right 12.1.1 The Equation of a Circle p.733-734
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The Equation of a Circle 11 (Page 734)
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arrow_right 12.1.2 Completing the Square for Equations of Circles p.738-741
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Completing the Square for Equations of Circles 24 (Page 739)
Completing the Square for Equations of Circles 25 (Page 739)
Completing the Square for Equations of Circles 26 (Page 739)
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Completing the Square for Equations of Circles 29 (Page 740)
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arrow_right 12.1.3 Introduction to Conic Sections p.744-745
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Introduction to Conic Sections 41 (Page 745)
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arrow_right 12.1.4 Graphing a Parabola Using The Focus and Directrix p.749-750
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Graphing a Parabola Using The Focus and Directrix 51 (Page 749)
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Graphing a Parabola Using The Focus and Directrix 53 (Page 750)
Graphing a Parabola Using The Focus and Directrix 54 (Page 750)
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