How would you find the center and radius of #x^2 + y^2 - 6x=0#?

1 Answer
Jun 26, 2018

Center: #(3,0)#
Radius: #3#

Explanation:

The standard form for the equation of a circle with center #(color(red)a,color(blue)b)# and radius #color(green)r# is
#color(white)("XXX")(x-color(red)a)^2+(y-color(blue)b)^2=color(green)r^2#

We need to convert the given equation #x^2+y^2-6x=0#
into this standard form.

Grouping the terms containing #x#
#x^2-6xcolor(white)("xxx")+color(white)("xxx")y^2color(white)("xxx")=0#

The #(y-color(blue)b)^2# term in the standard form can be written directly as #(y-color(blue)0)^2#
but the writing the #(x-color(red)a)^2# will require some manipulation of #x^2-6x#; to do this we will "complete the square:
#x^2-6color(magenta)(+3^2)color(white)("x")+color(white)("x")(y-color(blue)0)^2color(white)("xx")=0color(magenta)(+3^2)#

#(x-color(red)3)^2color(white)("xxx")+color(white)("x")(y-color(blue)0)^2color(white)("x")=color(green)3^2#
#color(white)("XXX")#which is the required form for a circle with
#color(white)("XXX")# center at #(color(red)3,color(blue)0)# and
#color(white)("XXX")# radius #color(green)3#