Exercise 5 Page 575

To find the equation of the circle we will use the Distance Formula. d = sqrt((x_2-x_1)^2 + (y_2-y_1)^2)In this form, the center of the circle is ( x_1, x_2) and d is the circle's radius. Let's label the radius r. r=sqrt((x_2- x_1)^2+(y_2- y_1)^2) We have been given that the circle has its center at ( 4, -1) and radius 3. Let's substitute these values into the formula and simplify.

r=sqrt((x_2- x_1)^2+(y_2- y_1)^2)

SubstituteValues

Substitute values

3=sqrt((x_2- 4)^2+(y_2-( - 1))^2)

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Simplify

RaiseEqn

LHS^2=RHS^2

3^2=(x_2-4)^2+(y_2-(- 1))^2

CalcPow

Calculate power

9=(x_2-4)^2+(y_2-(- 1))^2

NegNeg

- (- a)=a

9=(x_2-4)^2+(y_2+1)^2

RearrangeEqn

Rearrange equation

(x_2-4)^2+(y_2+1)^2=9

The equation of the circle can be written as (x_2-4)^2+(y_2+1)^2=9.