Question

for the equation $x^{2}+y^{2}-4x - 2y-20 = 0$, 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.
\[

$$\begin{align*} x^{2}+y^{2}-4x - 2y-20&=0\\ x^{2}-4x+y^{2}-2y&=20\\ (x - 2)^{2}-4+(y - 1)^{2}-1&=20\\ (x - 2)^{2}+(y - 1)^{2}&=20 + 4+1\\ (x - 2)^{2}+(y - 1)^{2}&=25 \end{align*}$$

\]

Step2: Identify 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 \((x - 2)^{2}+(y - 1)^{2}=25\), we have \(h = 2\), \(k = 1\), and \(r=\sqrt{25}=5\).

Step3: Find the \(x\) - intercepts

Set \(y = 0\) in the original equation \((x - 2)^{2}+(0 - 1)^{2}=25\).
\[

$$\begin{align*} (x - 2)^{2}+1&=25\\ (x - 2)^{2}&=24\\ x-2&=\pm\sqrt{24}=\pm2\sqrt{6}\\ x&=2\pm2\sqrt{6} \end{align*}$$

\]

Step4: Find the \(y\) - intercepts

Set \(x = 0\) in the original equation \((0 - 2)^{2}+(y - 1)^{2}=25\).
\[

$$\begin{align*} 4+(y - 1)^{2}&=25\\ (y - 1)^{2}&=21\\ y-1&=\pm\sqrt{21}\\ y&=1\pm\sqrt{21} \end{align*}$$

\]

Answer:

(a) Center \((h,k)=(2,1)\), radius \(r = 5\)
(b) To graph the circle, plot the center \((2,1)\) and then draw a circle with a radius of \(5\) units around it.
(c) \(x\) - intercepts: \(x=2 + 2\sqrt{6}\approx6.9\) and \(x=2-2\sqrt{6}\approx - 2.9\); \(y\) - intercepts: \(y=1+\sqrt{21}\approx5.6\) and \(y=1-\sqrt{21}\approx - 3.6\)