Question

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

Answer 1

We have a conic:

`x^2+y^2-4x+3 = 0`

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

`(x^2-4x) +(y^2) + 3 = 0`
`:. (x-2)^2-(2)^2 + y^2 + 3 = 0`

`:. (x-2)^2-4 + y^2 + 3 = 0`

`:. (x-2)^2 + y^2 = 1`

Comparing with:

`(x-a)^2+(y-b)^2 = r^2`

We see that the conic is a circle of radius `1` and centre `(2,0)`

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]}