# How do you graph x^2 + y^2 – 6x + 8y + 9 = 0?

Apr 18, 2016

Draw a circle with radius $4$ and center at $\left(3 , - 4\right)$

#### Explanation:

Given:
$\textcolor{w h i t e}{\text{XXX}} {x}^{2} + {y}^{2} - 6 x + 8 y + 9 = 0$

Re-arrange into standard circle equation form:

Re-group:
$\left({x}^{2} - 6 x\right) + \left({y}^{2} + 8 y\right) = - 9$

Complete the squares:
$\left({x}^{2} - 6 x \textcolor{red}{+ {3}^{2}}\right) + \left({y}^{2} + 8 y \textcolor{b l u e}{+ {4}^{2}}\right) = - 9 \textcolor{red}{+ {3}^{2}} \textcolor{b l u e}{+ {4}^{2}}$

Write as squared binomials and simplify the right side:
${\left(x - 3\right)}^{2} + {\left(y + 4\right)}^{2} = {4}^{2}$
or
(x-color(green)(3))^2+(y-color(green)(color(white)("")(-4)))^2=color(brown)(4^2)

Recalling that the standard circle equation is
color(white)("XXX")(x-color(green)(a))^2+(y-color(green)(b))^2=color(brown(r)^2
for a circle with center $\left(\textcolor{g r e e n}{a} , \textcolor{g r e e n}{b}\right)$ and radius $\textcolor{b r o w n}{r}$

The given equation is that of a circle with center (color(green)(3),color(green)(color(white)("")(-4))) and radius $\textcolor{b r o w n}{4}$

graph{x^2+y^2-6x+8y+9=0 [-5.72, 12.066, -8.194, 0.69]}