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.

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: Identify the center and radius

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

Step3: Find the x - intercepts

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

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

\]

Step4: Find the y - intercepts

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

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

\]

Answer:

(a) Center \((h,k)=(2,4)\), 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) x - intercepts: \(x=2 + 2\sqrt{5}\) and \(x=2-2\sqrt{5}\); y - intercepts: \(y=4 + 4\sqrt{2}\) and \(y=4-4\sqrt{2}\)