Core Connections Geometry, 2013
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Core Connections Geometry, 2013 View details
1. Section 12.1
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Exercise 24 Page 739

Practice makes perfect
a Let's start by recalling the equation of a circle in a graphing form.
(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.
(x-2)^2+(y+3)^2=25
â–Ľ
Rewrite
(x-2)^2+(y+3)^2=5^2

a=- (- a)

(x- 2)^2+(y-( -3))^2= 5^2
The center of the circle is the point ( 2, -3), and its radius is 5.
b Let's again recall the standard equation of a circle in a graphing form.
(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+2x+y^2+6y-6=0
â–Ľ
Rewrite
x^2+2x+y^2+6y+1+9 - 16=0
x^2+2x+1y^2+6y+9 - 16=0
x^2+2x+1+y^2+6y+9=16
x^2+2x+1^2+y^2+6y+3^2=4^2
x^2+2(x)(1)+1^2+y^2+2(y)(3)+3^2=4^2
(x+1)^2+(y+3)^2=4^2

a=- (- a)

(x-( -1))^2+(y-( -3))^2= 4^2
The center of the circle is the point ( -1, -3), and its radius is 4.