Question

for the equation x² + y² - 6x - 8y - 11 = 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 .
(type an ordered pair.)
the radius is r = .

(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: Rewrite the equation in standard form

The general equation of a circle is $(x - h)^2+(y - k)^2=r^2$, where $(h,k)$ is the center and $r$ is the radius. Given $x^{2}+y^{2}-6x - 8y-11 = 0$. We complete the square for $x$ and $y$ terms.
For the $x$ - terms: $x^{2}-6x=(x - 3)^{2}-9$. For the $y$ - terms: $y^{2}-8y=(y - 4)^{2}-16$.
So the equation becomes $(x - 3)^{2}-9+(y - 4)^{2}-16-11 = 0$, which simplifies to $(x - 3)^{2}+(y - 4)^{2}=36$.

Step2: Identify the center and radius

Comparing $(x - 3)^{2}+(y - 4)^{2}=36$ with $(x - h)^2+(y - k)^2=r^2$, we have $h = 3$, $k = 4$ and $r=\sqrt{36}=6$.

Step3: Find the $x$ - intercepts

Set $y = 0$ in the equation $(x - 3)^{2}+(y - 4)^{2}=36$. Then $(x - 3)^{2}+(0 - 4)^{2}=36$, i.e., $(x - 3)^{2}+16 = 36$, $(x - 3)^{2}=20$, $x-3=\pm\sqrt{20}=\pm2\sqrt{5}$, $x=3\pm2\sqrt{5}$. So the $x$ - intercepts are $(3 + 2\sqrt{5},0)$ and $(3 - 2\sqrt{5},0)$.

Step4: Find the $y$ - intercepts

Set $x = 0$ in the equation $(x - 3)^{2}+(y - 4)^{2}=36$. Then $(0 - 3)^{2}+(y - 4)^{2}=36$, $9+(y - 4)^{2}=36$, $(y - 4)^{2}=27$, $y - 4=\pm\sqrt{27}=\pm3\sqrt{3}$, $y=4\pm3\sqrt{3}$. So the $y$ - intercepts are $(0,4 + 3\sqrt{3})$ and $(0,4 - 3\sqrt{3})$.

Answer:

(a) The center is $(3,4)$. The radius is $r = 6$.
(c) A. The intercepts are $(3 + 2\sqrt{5},0),(3 - 2\sqrt{5},0),(0,4 + 3\sqrt{3}),(0,4 - 3\sqrt{3})$