# How do you find the center and radius of the circle x^2+y^2+8x+4y+16=0?

Sep 1, 2016

Center: color(green)(""(-4,-2))
Radius: $\textcolor{g r e e n}{2}$

#### Explanation:

If a circular equation is written in the form:
$\textcolor{w h i t e}{\text{XXX}} {\left(x - \textcolor{red}{a}\right)}^{2} + {\left(y - \textcolor{b l u e}{b}\right)}^{2} = {\textcolor{g r e e n}{r}}^{2}$
then it has a center at $\left(\textcolor{red}{a} , \textcolor{b l u e}{b}\right)$ and a radius of $\textcolor{g r e e n}{r}$

We will want to manipulate the given: ${x}^{2} + {y}^{2} + 8 x + 4 y + 16 = 0$
into this form.

First separating the $x$ terms, the $y$ terms and the constant as
$\textcolor{w h i t e}{\text{XXX}} \left(\textcolor{red}{{x}^{2} + 8 x}\right) + \left(\textcolor{b l u e}{{y}^{2} + 4 y}\right) = \textcolor{g r e e n}{- 16}$

Completing the square for each of the $x$ and $y$ sub-expressions:
$\textcolor{w h i t e}{\text{XXX}} \left(\textcolor{red}{{x}^{2} + 8 x + 16}\right) + \left(\textcolor{b l u e}{{y}^{2} + 4 y + 4}\right) = \textcolor{g r e e n}{- 16} \textcolor{red}{+ 16} \textcolor{b l u e}{+ 4}$

color(white)("XXX")(color(red)(x+4))^2+(color(blue)(y+2))^2= color(green)(4

$\textcolor{w h i t e}{\text{XXX}} {\left(x - \left(\textcolor{red}{- 4}\right)\right)}^{2} + {\left(y - \left(\textcolor{b l u e}{- 2}\right)\right)}^{2} = {\textcolor{g r e e n}{2}}^{2}$
with center at $\left(\textcolor{red}{- 4} , \textcolor{b l u e}{- 2}\right)$ and radius $\textcolor{g r e e n}{2}$

A graph of the original equation helps verify this result:
graph{x^2+y^2+8x+4y+16=0 [-9.184, 1.916, -4.89, 0.66]}