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: 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 \(r=\sqrt{36}=6\).

Step3: Graph the circle

The center of the circle is at the point \((2,4)\) and the radius is \(6\). Plot the center \((2,4)\) and then use the radius to find points on the circle by moving 6 units up, down, left, and right from the center.

Step4: Find the \(x\) - intercepts

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

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

\]

Step5: Find the \(y\) - intercepts

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

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

\]

Answer:

(a) Center \((h,k)=(2,4)\), radius \(r = 6\)
(b) Graph with center \((2,4)\) and radius \(6\)
(c) \(x\) - intercepts: \(x=2 + 2\sqrt{5}\) and \(x=2-2\sqrt{5}\); \(y\) - intercepts: \(y=4 + 4\sqrt{2}\) and \(y=4-4\sqrt{2}\)