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Calculus / Calculus: Early Transcendentals 1 / Chapter 10.2 / Problem 72E

Calculus: Early Transcendentals | 1st Edition | ISBN: 9780321570567 | Authors: William L. Briggs, Lyle Cochran, Bernard Gillett

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Solution: Equations of circles Use the results of Exercise

Chapter 8, Problem 72E

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QUESTION:

Use the results of Exercises 66-67 to describe and graph the following circles.

$r^2+2r(\cos \theta-3 \sin \theta)=4$

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QUESTION:

Use the results of Exercises 66-67 to describe and graph the following circles.

$r^2+2r(\cos \theta-3 \sin \theta)=4$

ANSWER:

Solution 72E

Step 1:

In this problem;

We have to describe and graph of the circle ${{r}^{2}+2\ r(cos(\theta{})-3sin(\theta{}))=4.}^{\ }$

We know that the polar equation ${{r}^{2}-2r(acos(\theta{})\ +bsin(\theta{}))={R}^{2}-{a}^{2}-{b}^{2}}^{\ }$ represents a circle of radius ‘R’ centered at ( a, b).

Consider , ${{r}^{2}\ +2\ r(cos(\theta{})\ -3sin(\theta{}))=4}^{\ }$

$\Rightarrow{}{r}^{2}-2r((-1)cos(\theta{})\ +3sin(\theta{})\ =4$

$\Rightarrow{}{r}^{2}-2r(\ (-1)cos(\theta{})\ +(3)sin(\theta{}))\ =4$

By , using the above result ${r}^{2}-2r(\ (-1)cos(\theta{})\ +(3)sin(\theta{}))\ =4$ represents a circle.

So , here a = -1, b = 3 , and ${R}^{2}-{a}^{2}-{b}^{2}$ = 4

$\Rightarrow{}{R}^{2}-(-1{)}^{2}-(-3{)}^{2}$ = 4

$\Rightarrow{}{R}^{2}-1-9\$ = 4

$\Rightarrow{}{R}^{2}-10$ = 4

$\Rightarrow{}{R}^{2}=10$ + 4 = 14

$\Rightarrow{}R\ =\sqrt{14}$ .

Therefore , ${r}^{2}+2\ r(cos(\theta{})-3sin(\theta{}))=4$

$\Rightarrow{}$ ${r}^{2}-2r(\ (-1)cos(\theta{})\ +(3)sin(\theta{}))\ =4$ represents a circle of radius ‘ $\sqrt{14}$ ’ centered at ( -1, 3).

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Solution: Equations of circles Use the results of Exercise

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Solution Found!

Solution: Equations of circles Use the results of Exercise

Questions & Answers

Study Tools You Might Need

Chapter: 9 Problem: 7.75

Chapter: 3 Problem: 16

Chapter: 27 Problem: 45

Chapter: 9 Problem: 3

Chapter: 33 Problem: 3

Chapter: 8 Problem: 4

Chapter: 10 Problem: 5

Chapter: 20 Problem: 30

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