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 + 7)^2 = 16
a) determine the center and radius of the circle.
the center of the circle is
(type an ordered pair.)
the radius is
(simplify your answer. type an exact answer, using radicals as needed.)
b) sketch the graph of the circle.
use the graphing tool to graph the circle.
click to enlarge graph
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.

Explanation:

Step1: Recall circle - standard form

The standard form of a circle equation is $(x - h)^2+(y - k)^2=r^2$, where $(h,k)$ is the center and $r$ is the radius.

Step2: Identify the center

For the equation $(x - 3)^2+(y+7)^2 = 16$, comparing with the standard - form, we have $h = 3$ and $k=-7$. So the center of the circle is $(3,-7)$.

Step3: Identify the radius

Since $r^2 = 16$, then $r=\sqrt{16}=4$.

Step4: Find the x - intercepts

Set $y = 0$ in the equation $(x - 3)^2+(y + 7)^2=16$. We get $(x - 3)^2+(0 + 7)^2=16$, i.e., $(x - 3)^2+49 = 16$, then $(x - 3)^2=16 - 49=-33$. Since the square of a real number cannot be negative, there are no x - intercepts.

Step5: Find the y - intercepts

Set $x = 0$ in the equation $(x - 3)^2+(y + 7)^2=16$. We have $(0 - 3)^2+(y + 7)^2=16$, i.e., $9+(y + 7)^2=16$, then $(y + 7)^2=16 - 9 = 7$. So $y+7=\pm\sqrt{7}$, and $y=-7\pm\sqrt{7}$. The y - intercepts are $(0,-7+\sqrt{7})$ and $(0,-7 - \sqrt{7})$.

Answer:

a) The center of the circle is $(3,-7)$. The radius is $4$.
b) (Sketching the circle requires a graph - drawing tool. The center is at the point $(3,-7)$ and the radius is 4 units. You can plot the center and then mark points 4 units away from the center in all directions to form the circle).
c) There are no x - intercepts. The y - intercepts are $(0,-7+\sqrt{7})$ and $(0,-7 - \sqrt{7})$.