# 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,

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I am not sure how you got $(x-2)^2$ in your solution - I think you need to look very carefully at the expansion of $(x-2)^2$ to see where you are going wrong. – Old John Jun 9 '12 at 21:47

When given an equation of the form $x^2-2x+y^2-2y+1=0$ the first step is to complete the square for $x$ and for $y$.

The idea is that if we have $x^2-2x$ we can write it as $(x+C)^2+D$ instead. Since know those that the coefficient of $x$ is $2C$, we know that $C=-1$, so we have: $$(x-1)^2=x^2-2x+1\implies x^2-2x = (x-1)^2-1$$

Therefore we can write it as $(x-1)^2-1$.

Similarly we can replace $y^2-2y$ by a similar term, so we have now:

\begin{align} &\underbrace{x^2-2x}+\underline{y^2-2y}+1 = 0 &&\text{complete the squares}\\ &\underbrace{(x-1)^2-1}+\underline{(y-1)^2-1}+1 =0 &&\text{sum the }1\text{'s}\\ &(x-1)^2+(y-1)^2-1=0 &&+1\\ &(x-1)^2+(y-1)^2=1 &&\text{ circle!} \end{align}

Therefore we have a circle of radius $1$ whose center is at $(1,1)$.

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This is how you do it without colors. – Asaf Karagila Jun 9 '12 at 23:26

Well it's a circle, I guess. Centre is $(1,1)$, radius is $1$.

If you want to draw it, you'll need this mighty instrument

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$$x^2+y^2-2x-2y+1=0\Longrightarrow (x-1)^2+(y-1)^2-2+1=0\Longrightarrow$$$$\Longrightarrow (x-1)^2+(y-1)^2=1$$Do you recognize it now?

Added In general we can complete the square as follows: $$ax^2+bx=a\left(x^2+\frac{b}{a}\right)=a\left(x+\frac{b}{2a}\right)^2-\frac{b^2}{4a}\,\,,\,\,a\neq 0$$

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yes,i got it! :) – mgh Jun 9 '12 at 21:51
@Meg Good! Just some corrections in your completing the squares, that's all. – DonAntonio Jun 9 '12 at 21:53

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?

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