Plotting  a quadratic equation in the $\,xy\,$- plane My question is:
Represent the following set of points in the $\,xy\,$- plane:
$$\left\{ (x,y)\,\, |\,\, x^2 + y^2 - 2x - 2y + 1 = 0 \right\}$$
What i got: $\,\,(x-2)^2 + (y-2)^2 = 1\,\,$
I am not getting what to do next.
Any help to solve this question would be greatly appreciated.
Thank you,

The first problem is that you carried out the algebra incorrectly. When you complete the square with $x^2-2x$ you should get $(x-1)^2-1$, which you can verify by multiplying it out. Similarly, $y^2-2y=(y-1)^2-1$. Thus, $$\begin{align*}x^2+y^2-2x-2y+1&=(x-1)^2-1+(y-1)^2-1+1\\
&=(x-1)^2+(y-1)^2-1\;,
\end{align*}$$
and the points where $x^2+y^2-2x-2y+1=0$ are the points where $(x-1)^2+(y-1)^2-1$, i.e., where $(x-1)^2+(y-1)^2=1$.
What’s the distance between the points $(x,y)$ and $(1,1)$? It’s $\sqrt{(x-1)^2+(y-1)^2}$, right? And if $(x-1)^2+(y-1)^2=1$, then $\sqrt{(x-1)^2+(y-1)^2}=\sqrt1=1$, so your set contains the points whose distance from $(1,1)$ is $1$. What does that set of points look like?
