Core Connections Integrated II, 2015
CC
Core Connections Integrated II, 2015 View details
Chapter Closure

Exercise 117 Page 592

(x-4)^2 + (y+2)^2 = 16

We want to write an equation for the graph from the following diagram.

The diagram shows a circle. Let's recall the standard equation of a circle. (x- h)^2 + (y- k)^2 = r^2 Here, ( h, k) is the center of the circle and r is the radius. To write an equation for our circle, we should identify its center and radius.

Center

To identify the center, let's connect the leftmost and the rightmost points of our circle with a line segment. Then, connect the points at the top and the bottom of the circle. The intersection of these segments is the center of our circle.

The center of our circle is the point ( 4, -2). To find the radius, let's find the distance from the center to a point on the circle. Here, let's use the leftmost point of the circle, (0,-2). Our center is 4 units to the right from this point, so the radius is 4 units. Now that we know the center and the radius, we will write the standard equation of our circle (x- h)^2 + (y- k)^2 = r^2 ⇓ (x- 4)^2 + ( y - ( -2))^2 = 4^2 Finally, let's simplify this equation.

(x-4)^2 + (y-(-2))^2 = 4^2
(x-4)^2 + (y+2)^2 = 4^2
(x-4)^2 + (y+2)^2 = 16