Core Connections Integrated II, 2015
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Core Connections Integrated II, 2015 View details
2. Section 10.2
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Exercise 68 Page 572

Practice makes perfect
a Like in Part A, 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 write the summation as subtraction of the opposite number.
(x+5)^2+y^2=10

a=- (- a)

(x-(-5))^2+y^2=10
(x-(-5))^2+(y-0)^2=10

a = ( sqrt(a) )^2

(x-(-5))^2+(y-0)^2=(sqrt(10))^2
The center of the circle is the point ( -5, 0), and its radius is sqrt(10).
b
Like in Part A. 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^2-6x+y^2-2y-5=0
â–Ľ
Rewrite
x^2-2(x)(3)+y^2-2(y)(1)-5=0
x^2-2(x)(3)+y^2-2(y)(1)+9+1-15=0
â–Ľ
Complete the square
x^2-2(x)(3)+9+y^2-2(y)(1)+1-15=0
x^2-2(x)(3)+3^2+y^2-2(y)(1)+1^2-15=0
(x-3)^2+(y-1)^2-15=0
(x-3)^2+(y-1)^2=15

a = ( sqrt(a) )^2

(x- 3)^2+(y- 1)^2=( sqrt(15))^2
The center of the circle is the point ( 3, 1), and its radius is sqrt(15).