# How do use the method of translation of axes to sketch the curve x^2+y^2+2x-6y+6=0?

Jan 24, 2017

See the method below

#### Explanation:

We rewrite the equation by completing the squares

$\left({x}^{2} + 2 x\right) + \left({y}^{2} - 6 y\right) = - 6$

$\left({x}^{2} + 2 x + 1\right) + \left({y}^{2} - 6 y + 9\right) = - 6 + 9 + 1$

${\left(x + 1\right)}^{2} + {\left(y - 3\right)}^{2} = 4$

This is a circle, center $\left(- 1 , 3\right)$ and radius $= 2$

We define new axes

$x ' = x + 1$

and $y ' = y - 3$

${\left(x '\right)}^{2} + {\left(y '\right)}^{2} = 4$

This is a circle, center $\left(0 , 0\right)$ and radius $= 2$

graph{((x+1)^2+(y-3)^2-4)(x^2+y^2-4)=0 [-10, 10, -5, 5]}