Answer
See below:
Work Step by Step
The general equation of a circle when it has a center other than the origin is $\left( x-{{h}^{2}} \right)+{{\left( y-k \right)}^{2}}={{r}^{2}}.$ Here, $\left( h,k \right)$ is the center of the circle and r is the radius of the circle.
Put $\left( h,k \right)=\left( -1,1 \right)$ and $r=1$ in the general equation of the circle:
$\begin{align}
& {{\left( x-1 \right)}^{2}}+{{\left( y-\left( -1 \right) \right)}^{2}}={{\left( 1 \right)}^{2}} \\
& {{\left( x-1 \right)}^{2}}+{{\left( y+1 \right)}^{2}}=1
\end{align}$
Now, simplify the above expression:
$\begin{align}
& {{\left( x-1 \right)}^{2}}+{{\left( y+1 \right)}^{2}}=1 \\
& {{x}^{2}}+1-2x+{{y}^{2}}+1+2y=1 \\
& {{x}^{2}}+{{y}^{2}}-2x+2y+2=1
\end{align}$
So, the center is $\left( 1,-2 \right)$ and radius is $1$.
The equation ${{\left( x-1 \right)}^{2}}+{{\left( y+1 \right)}^{2}}=1$ can be written as follows:
$y=\sqrt{1-{{\left( x-1 \right)}^{2}}}-1.$
Put $x=0$ in the above equation:
$\begin{align}
& y=\sqrt{1-{{\left( 0-1 \right)}^{2}}}-1 \\
& =\sqrt{0}-1 \\
& =-1
\end{align}$
Place $x=1$ ,
$\begin{align}
& y=\sqrt{1-{{\left( 1-1 \right)}^{2}}}-1 \\
& =\sqrt{1-0}-1 \\
& =\pm 1-1 \\
& =0,-2
\end{align}$
Place $x=2$ ,
$\begin{align}
& y=\sqrt{1-{{\left( 2-1 \right)}^{2}}}-1 \\
& =\sqrt{1-{{\left( 1 \right)}^{2}}}-1 \\
& =\sqrt{0}-1 \\
& =-1
\end{align}$
Thus, the points $\left( 0,-1 \right),\left( 1,0 \right),\left( 1,-2 \right),$ and $\left( 2,-1 \right)$ satisfy the equation:
$y=\sqrt{9-{{\left( x-1 \right)}^{2}}}-2.$
