given $x^2 + y^2 = 2x$. I want $(x-1)^2 + y^2 = 1$

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given $x^2 + y^2 = 2x$. I want $(x-1)^2 + y^2 = 1$

Is this completing the square?

my attempt:

$$x^2 + y^2 - 2x = 0$$

need $1$

$$x^2 + y^2 - 2x + 1 = 1$$

I forgot how to could someone explain please

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This is indeed completing the square. In this case, you have $$x^2+y^2=2x$$ Since we want to complete the square of the variable $x$, we will group the $x$ terms. So we get $$x^2-2x+y^2=0$$ Recall that $(a+b)^2=a^2+2ab+b^2$ and $(a-b)^2=a^2-2ab+b^2$. We want to complete the square using the $x$ terms we already have, namely $x^2-2x$, and get something of form $(a\pm b)^2$. In this case, since we have a minus sign in front of our second term, we deduce that we want to get something of form $(a-b)^2$, such that $$(a-b)^2=a^2-2ab+b^2=x^2-2x+b^2$$ for some $a,b$, which we have to figure out. Looking at the first term of this equality, we can clearly let $a=x$. Then, $2x=2ab=2xb$, so $b=1$. Then, $b^2=1$. But there is an issue here: here only have $x^2-2x+y^2=0$, and we do not have a $1$ any where. So we simply add to both sides. Our new equality is $$x^2-2x+1+y^2=1$$ Finally, using the fact that $(x-1)^2=x^2-2x+1$, we can conclude that $$(x-1)^2+y^2=1$$

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$$\begin{align}x^2 + y^2 - 2x + 1 = 1 &\leadsto x^2 - 2x + 1 + y^2 = 1 \\ &\leadsto (x - 1)^2 + y^2 = 1\end{align}$$

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$(x-1)^2 + y^2 = 1$

Expand to get:

$(x^2 -2x+1) + y^2=1$

Arrange the terms to get what you want:

$x^2 + y^2 = (2x - 1) + 1$

$x^2 + y^2 = 2x$