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.

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Accepted Answer

# 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 $y=1-\sqrt{21}\approx - 3.6$

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QUESTION IMAGE

Question

Explanation:

Step1: Rewrite the equation in standard form

Step2: Identify the center and radius

Step3: Find the \(x\) - intercepts

Step4: Find the \(y\) - intercepts

Answer: