# How do use the method of translation of axes to sketch the curve x^2+y^2-4x+3=0?

Jan 6, 2018

We have a conic:

${x}^{2} + {y}^{2} - 4 x + 3 = 0$

We rearrange and complete the square of the $x$ and $y$ terms independently:

$\left({x}^{2} - 4 x\right) + \left({y}^{2}\right) + 3 = 0$
$\therefore {\left(x - 2\right)}^{2} - {\left(2\right)}^{2} + {y}^{2} + 3 = 0$

$\therefore {\left(x - 2\right)}^{2} - 4 + {y}^{2} + 3 = 0$

$\therefore {\left(x - 2\right)}^{2} + {y}^{2} = 1$

Comparing with:

${\left(x - a\right)}^{2} + {\left(y - b\right)}^{2} = {r}^{2}$

We see that the conic is a circle of radius $1$ and centre $\left(2 , 0\right)$

Alternatively, if we compare to the origin centred circle ${x}^{2} + {y}^{2} = {r}^{2}$, we note that the conic is a circle of radius 1 translated $2$ units to the right.

So the graph is as follows:
graph{x^2+y^2-4x+3 = 0 [-3.656, 6.344, -2.5, 2.5]}