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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.2012-06-09

4 Answers 4

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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.2012-06-09
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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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    @Meg Good! Just some corrections in your completing the squares, that's all.2012-06-09
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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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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?