How would you find the center and radius of #x^2 + y^2 + 12x - 6y = 0#?

1 Answer
Steve M
Jun 1, 2018

An circle with radius #3sqrt(5)#, and centre #(-6,3)#

Explanation:

We have:

# x^2 + y^2 + 12x - 6y = 0 #

First we can collect "like" terms:

# {x^2 +12 x } + {y^2 - 6} = 0 #

Next we complete the square independently on the #x# and #y# terms

# :. {(x+6)^2-6^2} + {(y-3)^2-3^2} = 0 #

# :. (x+6)^2-36 + (y-3)^2-9 = 0 #

# :. (x+6)^2+ (y-3)^2 = 36+9 #

# :. (x+6)^2+ (y-3)^2 = 45 #

# :. (x+6)^2+ (y-3)^2 = (3sqrt(5))^2 #

And as such, we can identify the conic as an circle with radius #3sqrt(5)#, and centre #(-6,3)#

graph{x^2 + y^2 + 12x - 6y = 0 [-24.5, 15.5, -6.32, 13.68]}

Related questions
See all questions in Identify Critical Points
Impact of this question
5741 views around the world
You can reuse this answer
Creative Commons License