Exercise 9 Page 579

To begin, let's recall the Distance Formula. It is used to find the distance d between two points (x_1,y_1) and (x_2,y_2). d = sqrt((x_2-x_1)^2 + (y_2-y_1)^2) We will find the radius of our circle by substituting the given points into this formula and finding the distance between the center and the known point through which the circle passes.

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

SubstitutePoints

Substitute ( 0, 0) & ( 0, 6)

d=sqrt(( 0- 0)^2+( 6- 0)^2)

▼

Evaluate right-hand side

SubTerms

Subtract terms

d=sqrt(0^2+6^2)

CalcPow

Calculate power

d=sqrt(0+36)

IdPropAdd

Identity Property of Addition

d=sqrt(36)

CalcRoot

Calculate root

d=6

Let's now recall the standard equation of a circle. (x- h)^2+(y- k)^2= r^2 In this formula, ( h, k) is the center of the circle and r is its radius. We are told that the center of the circle is ( 0, 0). This information, together with r= 6, is enough to write the equation. (x- 0)^2+(y- 0)^2= 6^2 ⇕ x^2+y^2=36