Question

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?

Answer 1

`(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`