Question

How do you find the center and radius for `x^2+y^2-6x-6y+14=0`?

Answer 1

The center is `=(3,3)` and the radius is `=2`

Explanation:

The general equation of a circle, center `C=(a,b)` and radius `=r` is

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

Here, we have

`x^2+y^2-6x-6y+14=0`

Rearrange the equation and complete the square

`x^2-6x+y^2-6y=-14`

`x^2-6x+9+y^2-6y+9=-14+9+9`

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

The center is `=(3,3)` and the radius is `=2`

See the graph below.

graph{(x^2+y^2-6x-6y+14)=0 [-3.16, 12.64, -0.57, 7.33]}

Answer 2

`"centre "=(3,3)," radius"=2`

Explanation:

`"the equation of a circle in "color(blue)"standard form"` is.

`color(red)(bar(ul(|color(white)(2/2)color(black)((x-a)^2+(y-b)^2=r^2)color(white)(2/2)|)))`

`"where "(a,b)" are the coordinates of the centre and r"`
`"is the radius"`

`"to obtain this form "color(blue)"complete the square"`
`"on both the x and y terms"`

`x^2-6x+y^2-6x=-14`

`x^2+2(-3)x color(red)(+9)+y^2+2(-3)y color(magenta)(+9)=-14color(red)(+9)color(magenta)(+9)`

`(x-3)^2+(y-3)^2=4larrcolor(blue)"in standard form"`

`"centre "=(3,3)" and "r=sqrt4=2`