Question

for the equation x² + y² - 4x - 8y - 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. (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.)

Explanation:

Step1: Rewrite the equation in standard form

Complete the square for \(x\) and \(y\) terms.
\[

$$\begin{align*} x^{2}+y^{2}-4x - 8y-16&=0\\ x^{2}-4x+y^{2}-8y&=16\\ (x - 2)^{2}-4+(y - 4)^{2}-16&=16\\ (x - 2)^{2}+(y - 4)^{2}&=36 \end{align*}$$

\]

Step2: Find the center and radius

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.
For \((x - 2)^{2}+(y - 4)^{2}=36\), we have \(h = 2\), \(k = 4\), and \(r=\sqrt{36}=6\).
So the center \((h,k)=(2,4)\) and radius \(r = 6\).

Step3: Find the \(x\) - intercepts

Set \(y = 0\) in the equation \((x - 2)^{2}+(y - 4)^{2}=36\).
\[

$$\begin{align*} (x - 2)^{2}+(0 - 4)^{2}&=36\\ (x - 2)^{2}+16&=36\\ (x - 2)^{2}&=20\\ x-2&=\pm\sqrt{20}=\pm2\sqrt{5}\\ x&=2\pm2\sqrt{5} \end{align*}$$

\]
The \(x\) - intercepts are \((2 + 2\sqrt{5},0)\) and \((2-2\sqrt{5},0)\).

Step4: Find the \(y\) - intercepts

Set \(x = 0\) in the equation \((x - 2)^{2}+(y - 4)^{2}=36\).
\[

$$\begin{align*} (0 - 2)^{2}+(y - 4)^{2}&=36\\ 4+(y - 4)^{2}&=36\\ (y - 4)^{2}&=32\\ y-4&=\pm\sqrt{32}=\pm4\sqrt{2}\\ y&=4\pm4\sqrt{2} \end{align*}$$

\]
The \(y\) - intercepts are \((0,4 + 4\sqrt{2})\) and \((0,4-4\sqrt{2})\).

Answer:

(a) Center \((2,4)\), radius \(r = 6\)
(b) To graph the circle, plot the center \((2,4)\) and then draw a circle with radius \(r = 6\) around it.
(c) The intercepts are \((2 + 2\sqrt{5},0),(2-2\sqrt{5},0),(0,4 + 4\sqrt{2}),(0,4-4\sqrt{2})\)