Question

the equation of a circle is given in standard form. complete parts a through c to follow the step by step process to sketch the graph of the circle
$(x - 3)^2 + (y + 6)^2 = 16$

b) sketch the graph of the circle
use the graphing tool to graph the circle.

c) determine the x- and y-intercepts of the circle. select the correct choice and, if necessary, fill in the answer box within your choice
a. the x-intercept(s) is/are $x = \square$
(use a comma to separate answers as needed type an exact answer, using radicals as needed )
b. there is no x-intercept

Explanation:

Step1: Recall circle standard form

The standard form of a circle is $(x-h)^2+(y-k)^2=r^2$, where $(h,k)$ is the center and $r$ is the radius. For $(x-3)^2+(y+6)^2=16$, we have $h=3$, $k=-6$, $r=\sqrt{16}=4$.

Step2: Find x-intercepts (set $y=0$)

Substitute $y=0$ into the circle equation:
$$(x-3)^2+(0+6)^2=16$$
$$(x-3)^2+36=16$$
$$(x-3)^2=16-36=-20$$
Since a square cannot be negative, there are no real x-intercepts.

Step3: Find y-intercepts (set $x=0$)

Substitute $x=0$ into the circle equation:
$$(0-3)^2+(y+6)^2=16$$
$$9+(y+6)^2=16$$
$$(y+6)^2=16-9=7$$
$$y+6=\pm\sqrt{7}$$
$$y=-6\pm\sqrt{7}$$

Answer:

For part c) x-intercepts:

B. There is no x-intercept

For part c) y-intercepts (implied by the process):

$y=-6+\sqrt{7}, -6-\sqrt{7}$

For part b) graph:

The circle has center $(3, -6)$ and radius 4, matching the grid graph provided.