Question

What is the radius and centre of the circle: `x^2 + y^2 - 6x - 2y + 6 = 0`?

Answer 1

Radius is `2` and centre is `(3,1)`.

Explanation:

A general circle centred at `(a,b)` and with radius `r` has equation `(x-a)^2+(y-b)^2=r^2`.

So we need to manipulate the given equation to get it into that format, and we may do so by completing the square twice, once for `x` and once for `y` :

`x^2+y^2-6x-2y+6=0`

`therefore x^2-6x+((-6)/2)^2-((-6)/2)^2+y^2-2y+((-2)/2)^2-((-2)/2)^2+6=0`

`therefore(x-3)^2+(y-1)^2-9-1+6=0`

`therefore (x-3)^2+(y-1)^2=4`

Hence the radius is `r=sqrt4 = 2`, and the centre is `(3,1)`.