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

(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