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Core Connections Integrated II, 2015 View details

Chapter Closure

Exercise 116 Page 592

a We want to identify the center and the radius of the given circle based on its equation. Let's recall the graphing form of an equation of a circle.

(x- h)^2+(y- k)^2= r^2 In the equation above, ( h, k) is the center of a circle and r is its radius. Therefore, we should rewrite the given equation in this form so that we can identify them. x^2+&(y-3)^2=36 ⇕ (x- 0)^2+&(y- 3)^2= 6^2 Now we see that ( 0, 3) is the center and 6 is the radius of the given circle.

b Like in Part A, let's first rewrite the given equation so that it is in graphing form.

(x+1)^2+&(y+2)^2=8 ⇕ (x-( -1))^2+&(y-( -2))^2= sqrt(8)^2 Now we see that ( -1, -2) is the center and sqrt(8) is the radius of the given circle.

c Like in Parts A and B, we will first rewrite the given equation so that it is in graphing form. This time we will need to do it by completing the square.

x^2-8x+y^2+2y= -1

▼

Write in graphing form

SplitIntoFactors

Split into factors

x^2-2(4)x+y^2+2(1)y=-1

AddEqn

LHS+4^2=RHS+4^2

x^2-2(4)x+y^2+2(1)y+4^2=-1+4^2

AddEqn

LHS+1^2=RHS+1^2

x^2-2(4)x+y^2+2(1)y+4^2+1^2=-1+4^2+1^2