How do you convert this conic section to vertex/standard form #x^2 + y^2 + 6x - 16y +24#?

1 Answer
Dec 12, 2015

Answer:

This a circle with center #(-3,8)#, radius #7#, and standard form:
#color(white)("XXX")(x-(-3))^2+(y-8)^2=7^2#

Explanation:

Given:
#color(white)("XXX")x^2+y^2+6x-16y+24=0#
(actually, I assumed the #color(black)(=0)#; it wasn't "given")

Separating the #x# and #y# terms and moving the constant to the right side:
#color(white)("XXX")color(blue)(x^2+6x)+color(red)(y^2-16y)=color(green)(-24)#

Complete the squares for both the #x# and #y# sub-expressions:
#color(white)("XXX")color(blue)(x^2+6x+9)+color(red)(y^2-16y+64)=color(green)(-24)+color(blue)(9)+color(red)(64)#

Re-write as squared binomials; simplify the constant and write as a square:
#color(white)("XXX")color(blue)((x+3)^2)+color(red)((y-8)^2)=color(green)(7^2)#

Re-write #(x+3)# as #(x-(-3))#
#color(white)("XXX")color(blue)((x-(-3))^2) +color(red)((y-8)^2)= color(green)(7^2#

which is the same form as the general equation for a circle:
#color(white)("XXX")(x-color(blue)(a))^2+(y-color(red)(b))^2=color(green)(r)^2#
with center #(color(blue)(a),color(red)(b))#
and radius #color(green)(7)#

graph{x^2+y^2+6x-16y+24=0 [-14.57, 17.46, -0.53, 15.49]}