Algebra 2
Algebra 2
1st Edition
ISBN: 9780078884825
Author: McGraw-Hill/Glencoe
Publisher: Glencoe/McGraw-Hill School Pub Co
Question
Book Icon
Chapter 10.3, Problem 49PPS

(a)

To determine

To Find: The equation of a circle concentric with the circle at the right with the radius 4  units greater.

(a)

Expert Solution
Check Mark

Answer to Problem 49PPS

  (x+1)2+(y−4)2=36+165

Explanation of Solution

Given: AB is the diameter of the circle.

  Algebra 2, Chapter 10.3, Problem 49PPS , additional homework tip 1

Equation of the circle

  (x−h)2+(y−k)2=r2

Here

  A=(−5,2)B(3,6)

Thus

Find the center using the Mid-Point formula

  (h,k)=(x1+x22,y1+y22)

  x1=−5y1=2x2=3y2=6

Now

  (h,k)=(x1+x22,y1+y22)(h,k)=(−5+32,2+62)(h,k)=(−1,4)

Now

Find the radius using the distance formula

  r=(x2−x1)2+(y2−y1)2

Here

  x1=−12y1=16x2=−1y2=4

  d=(x2−x1)2+(y2−y1)2d=(3+5)2+(6−2)2d=64+16d=80d=45r=452r=25

Thus

Equation of the circle is

  (x+1)2+(y−4)2=(25)2(x+1)2+(y−4)2=80

Now the new center with the radius 4  units greater.

  r=25+4(h,k)=(−1,4)

Equation of the new circle is

  (x+1)2+(y−4)2=(25+4)2(x+1)2+(y−4)2=36+165

Hence, (x+1)2+(y−4)2=36+165

(b)

To determine

To Find: The equation of a circle concentric with the circle at the right with the radius 2  units less.

(b)

Expert Solution
Check Mark

Answer to Problem 49PPS

  (x+1)2+(y−4)2=24−85

Explanation of Solution

Given: AB is the diameter of the circle.

  Algebra 2, Chapter 10.3, Problem 49PPS , additional homework tip 2

Equation of the circle is

  (x+1)2+(y−4)2=(25)2(x+1)2+(y−4)2=80

Now the new center with the radius 2  units less

  r=25−2(h,k)=(−1,4)

Equation of the new circle is

  (x+1)2+(y−4)2=(25−2)2(x+1)2+(y−4)2=24−85

Hence, (x+1)2+(y−4)2=24−85

(c)

To determine

To Graph: The circles in part (a) and (b)

(c)

Expert Solution
Check Mark

Answer to Problem 49PPS

  Algebra 2, Chapter 10.3, Problem 49PPS , additional homework tip 3

Explanation of Solution

Given: AB is the diameter of the circle.

  Algebra 2, Chapter 10.3, Problem 49PPS , additional homework tip 4

Equation of the circle is

  (x+1)2+(y−4)2=80

Equation of the two concentric circle is

  (x+1)2+(y−4)2=36+165(x+1)2+(y−4)2=24−85

Thus

The graph

  Algebra 2, Chapter 10.3, Problem 49PPS , additional homework tip 5

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Chapter 10.3, Problem 48PPS
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Chapter 10.3, Problem 50PPS

Chapter 10 Solutions

Algebra 2

Ch. 10.1 - Prob. 1ACYPCh. 10.1 - Prob. 1BCYPCh. 10.1 - Prob. 2CYPCh. 10.1 - Prob. 3CYPCh. 10.1 - Prob. 1CYUCh. 10.1 - Prob. 2CYUCh. 10.1 - Prob. 3CYUCh. 10.1 - Prob. 4CYUCh. 10.1 - Prob. 5CYUCh. 10.1 - Prob. 6CYU
Ch. 10.1 - Prob. 7CYUCh. 10.1 - Prob. 8CYUCh. 10.1 - Prob. 9CYUCh. 10.1 - Prob. 10PPSCh. 10.1 - Prob. 11PPSCh. 10.1 - Prob. 12PPSCh. 10.1 - Prob. 13PPSCh. 10.1 - Prob. 14PPSCh. 10.1 - Prob. 15PPSCh. 10.1 - Prob. 16PPSCh. 10.1 - Prob. 17PPSCh. 10.1 - Prob. 18PPSCh. 10.1 - Prob. 19PPSCh. 10.1 - Prob. 20PPSCh. 10.1 - Prob. 21PPSCh. 10.1 - Prob. 22PPSCh. 10.1 - Prob. 23PPSCh. 10.1 - Prob. 24PPSCh. 10.1 - Prob. 25PPSCh. 10.1 - Prob. 26PPSCh. 10.1 - Prob. 27PPSCh. 10.1 - Prob. 28PPSCh. 10.1 - Prob. 29PPSCh. 10.1 - Prob. 30PPSCh. 10.1 - Prob. 31PPSCh. 10.1 - Prob. 32PPSCh. 10.1 - Prob. 33PPSCh. 10.1 - Prob. 34PPSCh. 10.1 - Prob. 35PPSCh. 10.1 - Prob. 36PPSCh. 10.1 - Prob. 37PPSCh. 10.1 - Prob. 38PPSCh. 10.1 - Prob. 39PPSCh. 10.1 - Prob. 40PPSCh. 10.1 - Prob. 41PPSCh. 10.1 - Prob. 42HPCh. 10.1 - Prob. 43HPCh. 10.1 - Prob. 44HPCh. 10.1 - Prob. 45HPCh. 10.1 - Prob. 46HPCh. 10.1 - Prob. 47STPCh. 10.1 - Prob. 48STPCh. 10.1 - Prob. 49STPCh. 10.1 - Prob. 50STPCh. 10.1 - Prob. 51SRCh. 10.1 - Prob. 52SRCh. 10.1 - 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Prob. 11SGRCh. 10 - Prob. 12SGRCh. 10 - Prob. 13SGRCh. 10 - Prob. 14SGRCh. 10 - Prob. 15SGRCh. 10 - Prob. 16SGRCh. 10 - Prob. 17SGRCh. 10 - Prob. 18SGRCh. 10 - Prob. 19SGRCh. 10 - Prob. 20SGRCh. 10 - Prob. 21SGRCh. 10 - Prob. 22SGRCh. 10 - Prob. 23SGRCh. 10 - Prob. 24SGRCh. 10 - Prob. 25SGRCh. 10 - Prob. 26SGRCh. 10 - Prob. 27SGRCh. 10 - Prob. 28SGRCh. 10 - Prob. 29SGRCh. 10 - Prob. 30SGRCh. 10 - Prob. 31SGRCh. 10 - Prob. 32SGRCh. 10 - Prob. 33SGRCh. 10 - Prob. 34SGRCh. 10 - Prob. 35SGRCh. 10 - Prob. 36SGRCh. 10 - Prob. 37SGRCh. 10 - Prob. 38SGRCh. 10 - Prob. 39SGRCh. 10 - Prob. 40SGRCh. 10 - Prob. 41SGRCh. 10 - Prob. 42SGRCh. 10 - Prob. 43SGRCh. 10 - Prob. 44SGRCh. 10 - Prob. 45SGRCh. 10 - Prob. 46SGRCh. 10 - Prob. 47SGRCh. 10 - Prob. 48SGRCh. 10 - Prob. 49SGRCh. 10 - Prob. 50SGRCh. 10 - Prob. 51SGRCh. 10 - Prob. 52SGRCh. 10 - Prob. 53SGRCh. 10 - Prob. 54SGRCh. 10 - Prob. 55SGRCh. 10 - Prob. 56SGRCh. 10 - Prob. 57SGRCh. 10 - Prob. 58SGRCh. 10 - Prob. 59SGRCh. 10 - Prob. 60SGRCh. 10 - Prob. 61SGRCh. 10 - Prob. 62SGRCh. 10 - Prob. 63SGRCh. 10 - Prob. 64SGRCh. 10 - Prob. 65SGRCh. 10 - Prob. 66SGRCh. 10 - Prob. 67SGRCh. 10 - Prob. 68SGRCh. 10 - Prob. 69SGRCh. 10 - Prob. 70SGRCh. 10 - Prob. 71SGRCh. 10 - Prob. 72SGRCh. 10 - Prob. 73SGRCh. 10 - Prob. 74SGRCh. 10 - Prob. 1PTCh. 10 - Prob. 2PTCh. 10 - Prob. 3PTCh. 10 - Prob. 4PTCh. 10 - Prob. 5PTCh. 10 - Prob. 6PTCh. 10 - Prob. 7PTCh. 10 - Prob. 8PTCh. 10 - Prob. 9PTCh. 10 - Prob. 10PTCh. 10 - Prob. 11PTCh. 10 - Prob. 12PTCh. 10 - Prob. 13PTCh. 10 - Prob. 14PTCh. 10 - Prob. 15PTCh. 10 - Prob. 16PTCh. 10 - Prob. 17PTCh. 10 - Prob. 18PTCh. 10 - Prob. 19PTCh. 10 - Prob. 20PTCh. 10 - Prob. 21PTCh. 10 - Prob. 22PTCh. 10 - Prob. 23PTCh. 10 - Prob. 24PTCh. 10 - Prob. 25PTCh. 10 - Prob. 1ECh. 10 - Prob. 2ECh. 10 - Prob. 3ECh. 10 - Prob. 4ECh. 10 - Prob. 5ECh. 10 - Prob. 1STPCh. 10 - Prob. 2STPCh. 10 - Prob. 3STPCh. 10 - Prob. 4STPCh. 10 - Prob. 5STPCh. 10 - Prob. 6STPCh. 10 - Prob. 7STPCh. 10 - Prob. 8STPCh. 10 - Prob. 9STPCh. 10 - Prob. 10STPCh. 10 - Prob. 11STPCh. 10 - Prob. 12STPCh. 10 - Prob. 13STPCh. 10 - Prob. 14STPCh. 10 - Prob. 15STP

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