How do you find the radius of a circle with the equation #x^2 + y^2 - 6x - 12y + 36 = 0#?

1 Answer
Alan P.
Dec 16, 2015

Convert the equation to the general form for a circle
#color(white)("XXX")(x-a)^2+(y-b)^2=r^2 to get the radius #r=3#

Explanation:

Given:
#color(white)("XXX")x^2+y^2-6x-12y+36=0#

Group #x# and #y# terms separately and transfer the constant to the right side
#color(white)("XXX")color(red)(x^2-6x)+color(blue)(y^2-12y)= -36#

Complete each of the #x# and #y# sub-expressions as squares
#color(white)("XXX")color(red)(x^2-6x+9)+color(blue)(y^2-12y+36) = -36 +color(red)(9)+color(blue)(36)#

Rewrite as a sum of squared binomials equal to a square
#color(white)("XXX")color(red)((x-3)^2) + color(blue)((y-6)^2) = 3^2#

This is the general form of a circle with center #(3,6)# and radius #3#
graph{x^2+y^2-6x-12y+36=0 [-3.07, 9.42, 2.885, 9.13]}

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