How do you find the center and radius of the circle #x^2 + y^2 -4x+12y-36 = 0#?

1 Answer
Sep 30, 2016

The center is #(2,-6)# and the radius is #2sqrt19#

Explanation:

#x^2+y^2-4x+12y-36=0#

Rearrange the terms by grouping the x terms, grouping the y terms, and moving the constant term to the other side
#x^2-4xcolor(white)(aaa)+y^2+12ycolor(white)(aaa)=36color(white)(aaa)#

Complete the square for x terms and the y terms.

To complete the square for the x terms, divide the coefficient of the "middle term" #color(red)4x# by 2, square the result to get #color(blue)4# and add #color(blue)4# to both sides of the equation.
#x^2-color(red)4x+color(blue)4color(white)(aaa)+y^2+12ycolor(white)(aa)=36+color(blue)4#

To complete the square for the y terms, divide the coefficient of the "middle term" #color(red)(12)y# by 2, square the result to get #color(blue)(36)# and add #color(blue)(36)# to both sides of the equation.
#x^2-4x+4color(white)(aaa)##+y^2+color(red)(12)y+color(blue)(36)=36+4+color(blue)(36)#

#x^2-4x+4color(white)(aaa)+y^2+12y+36=76#

Factor each group of terms.
#(x-2)(x-2)+(y+6)(y+6)=76#

#(x-2)^2+(y+6)^2=76#

This equation of a circle is #(x-h)^2 +(y-k)^2=r^2#
where #(h,k)# is the center and #r# is the radius.

Thus the center is #(2,-6)# and the radius is #sqrt76# or #2sqrt19#