Question

for the equation x² + y² - 8x - 4y - 16 = 0, do the following. (a) find the center (h,k) and radius r of the circle. (b) graph the circle. (c) find the intercepts, if any. (a) the center is (4,2). (type an ordered pair.) the radius is r = 6. (b) use the graphing tool to graph the circle. (c) find the intercepts, if any. select the correct choice and, if necessary, fill in the answer box to complete your choice. a. the intercept(s) is/are . (type an ordered pair. use a comma to separate answers as needed. type exact answers for each coordinate, using radicals as needed.) b. there is no intercept.

Explanation:

Step1: Recall the standard - form of a circle equation

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. Given the equation $x^{2}+y^{2}-8x - 4y-16 = 0$, we complete the square for $x$ and $y$ terms.
For the $x$ - terms: $x^{2}-8x=(x - 4)^2-16$. For the $y$ - terms: $y^{2}-4y=(y - 2)^2-4$.
So the equation becomes $(x - 4)^2-16+(y - 2)^2-4-16 = 0$, which simplifies to $(x - 4)^2+(y - 2)^2=36$.

Step2: Find the $x$ - intercepts

To find the $x$ - intercepts, set $y = 0$ in the equation $(x - 4)^2+(y - 2)^2=36$.
We get $(x - 4)^2+(0 - 2)^2=36$, i.e., $(x - 4)^2+4 = 36$, then $(x - 4)^2=32$.
Taking the square root of both sides, $x-4=\pm\sqrt{32}=\pm4\sqrt{2}$. So $x = 4\pm4\sqrt{2}$. The $x$ - intercepts are $(4 + 4\sqrt{2},0)$ and $(4 - 4\sqrt{2},0)$.

Step3: Find the $y$ - intercepts

To find the $y$ - intercepts, set $x = 0$ in the equation $(x - 4)^2+(y - 2)^2=36$.
We have $(0 - 4)^2+(y - 2)^2=36$, i.e., $16+(y - 2)^2=36$, then $(y - 2)^2=20$.
Taking the square root of both sides, $y - 2=\pm\sqrt{20}=\pm2\sqrt{5}$. So $y=2\pm2\sqrt{5}$. The $y$ - intercepts are $(0,2 + 2\sqrt{5})$ and $(0,2 - 2\sqrt{5})$.

Answer:

A. $(4 + 4\sqrt{2},0),(4 - 4\sqrt{2},0),(0,2 + 2\sqrt{5}),(0,2 - 2\sqrt{5})$