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

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:
$\textcolor{w h i t e}{\text{XXX}} {x}^{2} + {y}^{2} - 6 x - 12 y + 36 = 0$

Group $x$ and $y$ terms separately and transfer the constant to the right side
$\textcolor{w h i t e}{\text{XXX}} \textcolor{red}{{x}^{2} - 6 x} + \textcolor{b l u e}{{y}^{2} - 12 y} = - 36$

Complete each of the $x$ and $y$ sub-expressions as squares
$\textcolor{w h i t e}{\text{XXX}} \textcolor{red}{{x}^{2} - 6 x + 9} + \textcolor{b l u e}{{y}^{2} - 12 y + 36} = - 36 + \textcolor{red}{9} + \textcolor{b l u e}{36}$

Rewrite as a sum of squared binomials equal to a square
$\textcolor{w h i t e}{\text{XXX}} \textcolor{red}{{\left(x - 3\right)}^{2}} + \textcolor{b l u e}{{\left(y - 6\right)}^{2}} = {3}^{2}$

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