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 - 1)^2+(y + 7)^2 = 4
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
a. the x - intercept(s) is/are x=
(use a comma to separate answers as needed. type an exact answer, using radicals as needed )
b. there is no x - intercept
select the correct choice and, if necessary, fill in the answer box within your choice
a. the y - intercept(s) is/are y=
(use a comma to separate answers as needed. type an exact answer, using radicals as needed )
b. there is no y - intercept

Explanation:

Step1: Recall the standard - form of a circle

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 the circle $(x - 1)^2+(y + 7)^2 = 4$, the center is $(1,-7)$ and the radius $r = 2$.

Step2: Find the x - intercepts

Set $y = 0$ in the equation of the circle:
$(x - 1)^2+(0 + 7)^2=4$, which simplifies to $(x - 1)^2+49 = 4$, then $(x - 1)^2=-45$. Since the square of a real number cannot be negative, there are no real - valued x - intercepts.

Step3: Find the y - intercepts

Set $x = 0$ in the equation of the circle:
$(0 - 1)^2+(y + 7)^2=4$, which is $1+(y + 7)^2=4$. Then $(y + 7)^2=3$. Taking the square root of both sides, we get $y+7=\pm\sqrt{3}$. So $y=-7\pm\sqrt{3}$.

Answer:

B. There is no x - intercept
A. The y - intercept(s) is/are $y=-7+\sqrt{3},-7 - \sqrt{3}$