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

##### 2 Answers

#### Answer:

The center is

#### Explanation:

The general equation of a circle, center

Here, we have

Rearrange the equation and complete the square

The center is

See the graph below.

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

#### Answer:

#### 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#