# What is the center and radius of the circle with equation x^2 + y^2 + 6x + 16y + 64 = 0?

May 4, 2016

#### Answer:

center: $\left(- 3 , - 8\right)$
radius: $3$

#### Explanation:

Recall that the general equation of a circle is:

$\textcolor{b l u e}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} {\left(x - h\right)}^{2} + {\left(y - k\right)}^{2} = {r}^{2} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

where:
$x =$x-coordinate
$h =$x-coordinate of circle's centre
$y =$y-coordinate
$k =$y-coordinate of circle's centre
$r =$radius of circle

Given the equation,

${x}^{2} + {y}^{2} + 6 x + 16 y + 64 = 0$

Subtract $64$ from both sides.

${x}^{2} + {y}^{2} + 6 x + 16 y = - 64$

Grouping and bracketing the terms with $x$ and $y$ separately,

${x}^{2} + 6 x + {y}^{2} + 16 y = - 64$

$\left({x}^{2} + 6 x\right) + \left({y}^{2} + 16 y\right) = - 64$

Complete the square within each bracket.

$\left({x}^{2} + 6 x \textcolor{w h i t e}{i} \textcolor{red}{+ {\left(\frac{6}{2}\right)}^{2}}\right) + \left({y}^{2} + 16 y \textcolor{w h i t e}{i} \textcolor{red}{+ {\left(\frac{16}{2}\right)}^{2}}\right) = - 64$

$\left({x}^{2} + 6 x \textcolor{w h i t e}{i} \textcolor{red}{+ 9}\right) + \left({y}^{2} + 16 y \textcolor{w h i t e}{i} \textcolor{red}{+ 64}\right) = - 64$

Add $9$ and $64$ to the right side of the equation.

$\left({x}^{2} + 6 x \textcolor{w h i t e}{i} \textcolor{\mathrm{da} r k \mathmr{and} a n \ge}{+ 9}\right) + \left({y}^{2} + 16 y \textcolor{w h i t e}{i} \textcolor{\mathrm{da} r k \mathmr{and} a n \ge}{+ 64}\right) = - 64 \textcolor{w h i t e}{i} \textcolor{\mathrm{da} r k \mathmr{and} a n \ge}{+ 64} \textcolor{w h i t e}{i} \textcolor{\mathrm{da} r k \mathmr{and} a n \ge}{+ 9}$

$\left({x}^{2} + 6 x + 9\right) + \left({y}^{2} + 16 y + 64\right) = 9$

Rewrite each bracketed expression in factored form.

${\left(x + 3\right)}^{2} + {\left(y + 8\right)}^{2} = 9$

According to the general equation of a circle, the centre and radius can be determined by looking at the equation for the circle as,

${\left(x - \left(\textcolor{red}{- 3}\right)\right)}^{2} + {\left(y - \left(\textcolor{b l u e}{- 8}\right)\right)}^{2} = {\textcolor{p u r p \le}{3}}^{2}$

$\underline{\text{Reminder} : {\left(x - \textcolor{red}{h}\right)}^{2} + {\left(y - \textcolor{b l u e}{k}\right)}^{2} = {\textcolor{p u r p \le}{r}}^{2}}$

Since the centre is $\left(\textcolor{red}{h} , \textcolor{b l u e}{k}\right)$ and the radius is $\textcolor{p u r p \le}{r}$, using the derived equation, we can determine that the centre of the circle is $\left(\textcolor{red}{- 3} , \textcolor{b l u e}{- 8}\right)$ and the radius is $\textcolor{p u r p \le}{3}$.