Answer
The center and radius of a circle from its equation ${{x}^{2}}+{{y}^{2}}-4x+2y-4=0$ is $\left( 2,-1 \right)$ and 3 respectively. The domain of the graph of the circle is $\left[ -1,5 \right]$, and the range of the graph of the circle is $\left[ -4,2 \right]$.
Work Step by Step
It is known that he standard equation of a circle that has radius $r$ and center $\left( h,k \right)$ is given by:
${{\left( x-h \right)}^{2}}+{{\left( y-k \right)}^{2}}={{r}^{2}}$ (I)
Rewrite the equation in the form of the standard equation of a circle as shown below:
Rearrange the given equation and factorize:
$\begin{align}
& {{x}^{2}}-4x+{{y}^{2}}+2y-4=0 \\
& {{x}^{2}}-4x+-2+{{y}^{2}}+2y-2=0 \\
& {{x}^{2}}-4x+4+{{y}^{2}}+2y+1-4-1-4=0
\end{align}$
Use the identity ${{a}^{2}}-2ab+{{b}^{2}}={{\left( a-b \right)}^{2}}$ to factorize the equation:
$\begin{align}
& {{x}^{2}}-4x+4+{{y}^{2}}+2y+1=4+1+4 \\
& {{\left( x-2 \right)}^{2}}+{{\left( y+1 \right)}^{2}}=9
\end{align}$
${{\left( x-2 \right)}^{2}}+{{\left( y-\left( -1 \right) \right)}^{2}}={{3}^{2}}$ (II)
Compare equations (I) and (II) to find the value of h, k and r:
Here, $h=2,k=-1,r=3$
So, the center of the circle will be $\left( h,k \right)=\left( 2,-1 \right)$ and its radius will be $r=3$.
It can be seen from the graph that the $x$ values can take values from –1 to 5 and $y$ values are also within the range –4 to 2.
Hence, the domain and range of this circle are $\left[ -1,5 \right]\text{ and }\left[ -4,2 \right]$, respectively.
Hence, the coordinate for the center of the circle is $\left( 2,-1 \right)$ and its radius is 3. The domain is $\left[ -1,5 \right]$ and the range is $\left[ -4,2 \right]$.
