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^{2}+(y - 2)^{2}=4) a) determine the center and radius of the circle. the center of the circle is (square) (type an ordered pair.) the radius is (square) (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. 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 is $(x - h)^2+(y - k)^2=r^2$, where $(h,k)$ is the center and $r$ is the radius.
For the equation $x^{2}+(y - 2)^{2}=4$, we have $h = 0$, $k = 2$, and $r^{2}=4$.

Step2: Find the center

Comparing with the standard - form, the center of the circle $(h,k)$ is $(0,2)$.

Step3: Find the radius

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

Step4: Find the x - intercepts

Set $y = 0$ in the equation $x^{2}+(y - 2)^{2}=4$.
We get $x^{2}+(0 - 2)^{2}=4$, which simplifies to $x^{2}+4 = 4$, then $x^{2}=0$, so $x = 0$. The x - intercept is $(0,0)$.

Step5: Find the y - intercepts

Set $x = 0$ in the equation $x^{2}+(y - 2)^{2}=4$.
We have $(0)^{2}+(y - 2)^{2}=4$, so $(y - 2)^{2}=4$.
Taking the square root of both sides: $y−2=\pm2$.
When $y−2 = 2$, $y = 4$; when $y−2=-2$, $y = 0$. The y - intercepts are $(0,0)$ and $(0,4)$.

Answer:

a) The center of the circle is $(0,2)$. The radius is $2$.
b) To sketch the circle: Plot the center $(0,2)$. Then, from the center, move 2 units up, down, left, and right to get four points on the circle. Connect these points to form a circle.
c) The x - intercept is $(0,0)$. The y - intercepts are $(0,0)$ and $(0,4)$.