Question

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

Answer 1

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]}