How do you find the center and radius of the circle #4x^2 - 16x + 4y^2 - 24y + 36 = 0#?

1 Answer
Shwetank Mauria
Jun 15, 2016

#4x^2-16x+4y^2-24y+36=0# is a circle with center at #(2,3)# and radius #2#.

Explanation:

Dividing each term by #4# in #4x^2-16x+4y^2-24y+36=0#, we get

#x^2-4x+y^2-6y+9=0#

Now for converting #x^2-4x# and #y^2-6y# into complete squares, we need to add #4# and #9# respectively. As #9# is already there, adding #4# on both sides, we get

#x^2-4x+4+y^2-6y+9=4#

or #(x-2)^2+(y-3)^2=2^2#

This is the equation of a circle with center at #(2,3)# and radius #2#. Its graph is as given below.
graph{4x^2-16x+4y^2-24y+36=0 [-8.46, 11.54, -2.28, 7.72]}

Related questions
See all questions in Identify Critical Points
Impact of this question
14638 views around the world
You can reuse this answer
Creative Commons License