Question

for the equation x² + y² - 2x - 6y - 26 = 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 below 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 $(x - h)^2+(y - k)^2=r^2$. Given $x^{2}+y^{2}-2x - 6y-26 = 0$. Group the $x$ - terms and $y$ - terms: $(x^{2}-2x)+(y^{2}-6y)=26$. Complete the square for the $x$ - terms: $x^{2}-2x=(x - 1)^{2}-1$. Complete the square for the $y$ - terms: $y^{2}-6y=(y - 3)^{2}-9$. Then $(x - 1)^{2}-1+(y - 3)^{2}-9 = 26$, which simplifies to $(x - 1)^{2}+(y - 3)^{2}=36$.

Step2: Identify the center and radius

Comparing $(x - 1)^{2}+(y - 3)^{2}=36$ with $(x - h)^2+(y - k)^2=r^2$, we have $h = 1$, $k = 3$, and $r=\sqrt{36}=6$. So the center is $(1,3)$ and the radius $r = 6$.

Step3: Find the $x$ - intercepts

Set $y = 0$ in the equation $(x - 1)^{2}+(y - 3)^{2}=36$. Then $(x - 1)^{2}+(0 - 3)^{2}=36$, i.e., $(x - 1)^{2}+9 = 36$, $(x - 1)^{2}=27$, $x-1=\pm3\sqrt{3}$, $x=1\pm3\sqrt{3}$. The $x$ - intercepts are $(1 + 3\sqrt{3},0)$ and $(1-3\sqrt{3},0)$.

Step4: Find the $y$ - intercepts

Set $x = 0$ in the equation $(x - 1)^{2}+(y - 3)^{2}=36$. Then $(0 - 1)^{2}+(y - 3)^{2}=36$, i.e., $1+(y - 3)^{2}=36$, $(y - 3)^{2}=35$, $y-3=\pm\sqrt{35}$, $y=3\pm\sqrt{35}$. The $y$ - intercepts are $(0,3+\sqrt{35})$ and $(0,3 - \sqrt{35})$.

Answer:

(a) The center is $(1,3)$; The radius is $r = 6$
(c) A. The intercept(s) is/are $(1 + 3\sqrt{3},0),(1-3\sqrt{3},0),(0,3+\sqrt{35}),(0,3 - \sqrt{35})$