What is the standard form of the equation of a circle with centre and radius of the circle x^2 + y^2 - 4x + 8y - 80?

1 Answer
Jan 8, 2016

Answer:

#(x-2)^2+(y-(-4))^2=10^2#

Explanation:

The general standard form for the equation of a circle is
#color(white)("XXX")(x-a)^2+(y-b)^2=r^2#
for a circle with center #(a,b)# and radius #r#

Given
#color(white)("XXX")x^2+y^2-4x+8y-80 (=0)color(white)("XX")#(note: I added the #=0# for the question to make sense).

We can transform this into the standard form by the following steps:

Move the #color(orange)("constant")# to the right side and group the #color(blue)(x)# and #color(red)(y)# terms separately on the left.
#color(white)("XXX")color(blue)(x^2-4x)+color(red)(y^2+8y)=color(orange)(80)#

Complete the square for each of the #color(blue)(x)# and #color(red)(y)# sub-expressions.
#color(white)("XXX")color(blue)(x^2-4x+4)+color(red)(y^2+8y+16)=color(orange)(80)color(blue)(+4)color(red)(+16)#

Re-write the #color(blue)(x)# and #color(red)(y)# sub-expressions as binomial squares and the constant as a square.
#color(white)("XXX")color(blue)((x-2)^2)+color(red)((y+4)^2) = color(green)(10^2)#

Often we would leave it in this form as "good enough",
but technically this wouldn't make the #y# sub-expression into the form #(y-b)^2# (and might cause confusion as to the y component of the center coordinate).

So more accurately:
#color(white)("XXX")color(blue)((x-2)^2)+color(red)((y-(-4))^2=color(green)(10^2)#
with center at #(2,-4)# and radius #10#