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

2 Answers
Narad T.
Aug 7, 2018

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]}

Jim G.
Aug 7, 2018

#"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#

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