# How do you convert this conic section to vertex/standard form x^2 + y^2 + 6x - 16y +24?

Dec 12, 2015

This a circle with center $\left(- 3 , 8\right)$, radius $7$, and standard form:
$\textcolor{w h i t e}{\text{XXX}} {\left(x - \left(- 3\right)\right)}^{2} + {\left(y - 8\right)}^{2} = {7}^{2}$

#### Explanation:

Given:
$\textcolor{w h i t e}{\text{XXX}} {x}^{2} + {y}^{2} + 6 x - 16 y + 24 = 0$
(actually, I assumed the $\textcolor{b l a c k}{= 0}$; it wasn't "given")

Separating the $x$ and $y$ terms and moving the constant to the right side:
$\textcolor{w h i t e}{\text{XXX}} \textcolor{b l u e}{{x}^{2} + 6 x} + \textcolor{red}{{y}^{2} - 16 y} = \textcolor{g r e e n}{- 24}$

Complete the squares for both the $x$ and $y$ sub-expressions:
$\textcolor{w h i t e}{\text{XXX}} \textcolor{b l u e}{{x}^{2} + 6 x + 9} + \textcolor{red}{{y}^{2} - 16 y + 64} = \textcolor{g r e e n}{- 24} + \textcolor{b l u e}{9} + \textcolor{red}{64}$

Re-write as squared binomials; simplify the constant and write as a square:
$\textcolor{w h i t e}{\text{XXX}} \textcolor{b l u e}{{\left(x + 3\right)}^{2}} + \textcolor{red}{{\left(y - 8\right)}^{2}} = \textcolor{g r e e n}{{7}^{2}}$

Re-write $\left(x + 3\right)$ as $\left(x - \left(- 3\right)\right)$
color(white)("XXX")color(blue)((x-(-3))^2) +color(red)((y-8)^2)= color(green)(7^2

which is the same form as the general equation for a circle:
$\textcolor{w h i t e}{\text{XXX}} {\left(x - \textcolor{b l u e}{a}\right)}^{2} + {\left(y - \textcolor{red}{b}\right)}^{2} = {\textcolor{g r e e n}{r}}^{2}$
with center $\left(\textcolor{b l u e}{a} , \textcolor{red}{b}\right)$
and radius $\textcolor{g r e e n}{7}$

graph{x^2+y^2+6x-16y+24=0 [-14.57, 17.46, -0.53, 15.49]}