Exercise 6 Page 778

Let's consider the given circle.

Now, 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 ( 3,- 4) & ( 0,- 6)

d=sqrt(( 0- 3)^2+( - 6-( - 4))^2)

▼

Evaluate right-hand side

SubNeg

a-(- b)=a+b

d=sqrt((0-3)^2+(- 6+4)^2)

AddSubTerms

Add and subtract terms

d=sqrt((- 3)^2+(- 2)^2)

CalcPow

Calculate power

d=sqrt(9+4)

AddTerms

Add terms

d=sqrt(13)

Let's now recall the standard form of an 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. As we can see in the diagram, the center of the circle is ( 3, - 4). This information, together with r= sqrt(13), is enough to write the equation. (x- 3)^2+(y-( - 4))^2=( sqrt(13))^2 ⇕ (x-3)^2+(y+4)^2=13