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Big Ideas Math Geometry, 2014

BI

Big Ideas Math Geometry, 2014 View details

Cumulative Assessment

Exercise 4 Page 588

Adapt the given equation to match the standard equation of a circle.

B

Practice makes perfect

Let's start by recalling the standard equation of a circle. (x- h)^2+(y- k)^2= r^2 Here, the center is the point ( h, k) and the radius is r. We will rewrite the given equation to match this form, and then we can identify the center and the radius. In this case we will need to complete the square twice — once for each variable.

x^2+y^2+14x-16y+77=0

LHS+36=RHS+36

x^2+y^2+14x-16y+113=36

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Rewrite

Rewrite 113 as 49+64

x^2+y^2+14x-16y+49+64=36

CommutativePropAdd

Commutative Property of Addition

x^2+14x+49+y^2-16y+64=36

Write as a power

x^2+14x+7^2+y^2-16y+8^2=36

SplitIntoFactors

Split into factors

x^2+2(x)(7)+7^2+y^2-2(y)(8)+8^2=36

a^2± 2ab+b^2=(a± b)^2

(x+7)^2+(y-8)^2=36

a = ( sqrt(a) )^2

(x+7)^2+(y-8)^2=(sqrt(36))^2

Calculate root

(x+ 7)^2+(y- 8)^2=( 6)^2

The center of the circle is the point ( - 7, 8), and its radius is 6. Therefore choice B is the correct answer.

Subchapter links

Exercises p.588-589