Core Connections Integrated II, 2015
CC
Core Connections Integrated II, 2015 View details
1. Section 10.1
Continue to next subchapter

Exercise 7 Page 549

Practice makes perfect
a We want to write the equation of the given circle. To do so, let's first recall the equation of a circle in graphing form.
(x- h)^2+(y- k)^2= r^2 In the equation above, ( h, k) is the center of the circle and r is its radius. We need to identify all of them to write our equation. Let's take a look at the given diagram.

It seems that the origin is the center of a given circle and points (3,0), (-3,0), (0,3), and (0,-3) all lie on it.

Since the center is ( 0, 0), we can partially complete the equation of a circle. (x- 0)^2+(y- 0)^2=r^2 ⇕ x^2+y^2 = r^2 Since (x, y) are the coordinates of any point lying on a circle, we can choose any of the points we found and substitute it into the equation above. That way we will get an equation which we can solve for r. Let's choose the point (3,0).
x^2+y^2 = r^2
3^2+ 0^2=r^2
â–Ľ
Solve for r
9+0=r^2
9=r^2
±sqrt(9)=r
±3 = r
r = ± 3

r > 0

r = 3
Now we can complete our equation. x^2+y^2= 3^2 ⇕ x^2+y^2=9
b To sketch the given equation, we should first rewrite it so that it matches the equation of a circle in a graphing form.

x^2+y^2=81 ⇕ (x- 0)^2+(y- 0)^2 = 9^2 As we can see, point ( 0, 0) is the center and 9 is the radius of the given circle. Therefore, to draw this circle we connect all points that are 9 units away from the origin.