# 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?

Jan 8, 2016

${\left(x - 2\right)}^{2} + {\left(y - \left(- 4\right)\right)}^{2} = {10}^{2}$

#### Explanation:

The general standard form for the equation of a circle is
$\textcolor{w h i t e}{\text{XXX}} {\left(x - a\right)}^{2} + {\left(y - b\right)}^{2} = {r}^{2}$
for a circle with center $\left(a , b\right)$ and radius $r$

Given
$\textcolor{w h i t e}{\text{XXX")x^2+y^2-4x+8y-80 (=0)color(white)("XX}}$(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 $\textcolor{\mathmr{and} a n \ge}{\text{constant}}$ to the right side and group the $\textcolor{b l u e}{x}$ and $\textcolor{red}{y}$ terms separately on the left.
$\textcolor{w h i t e}{\text{XXX}} \textcolor{b l u e}{{x}^{2} - 4 x} + \textcolor{red}{{y}^{2} + 8 y} = \textcolor{\mathmr{and} a n \ge}{80}$

Complete the square for each of the $\textcolor{b l u e}{x}$ and $\textcolor{red}{y}$ sub-expressions.
$\textcolor{w h i t e}{\text{XXX}} \textcolor{b l u e}{{x}^{2} - 4 x + 4} + \textcolor{red}{{y}^{2} + 8 y + 16} = \textcolor{\mathmr{and} a n \ge}{80} \textcolor{b l u e}{+ 4} \textcolor{red}{+ 16}$

Re-write the $\textcolor{b l u e}{x}$ and $\textcolor{red}{y}$ sub-expressions as binomial squares and the constant as a square.
$\textcolor{w h i t e}{\text{XXX}} \textcolor{b l u e}{{\left(x - 2\right)}^{2}} + \textcolor{red}{{\left(y + 4\right)}^{2}} = \textcolor{g r e e n}{{10}^{2}}$

Often we would leave it in this form as "good enough",
but technically this wouldn't make the $y$ sub-expression into the form ${\left(y - b\right)}^{2}$ (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 $\left(2 , - 4\right)$ and radius $10$