# How do you write an equation for a circle with center (-8, 13) and radius 8?

Apr 26, 2016

${\left(x + 8\right)}^{2} + {\left(y - 13\right)}^{2} = 64$

#### Explanation:

The standard form of the equation of a circle is.

$\textcolor{red}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{{\left(x - a\right)}^{2} + {\left(y - b\right)}^{2} = {r}^{2}} \textcolor{w h i t e}{\frac{a}{a}} |}}}$
where (a ,b) are the coords of the centre and r,the radius

here a = -8 , b = 13 and r = 8

$\Rightarrow {\left(x + 8\right)}^{2} + {\left(y - 13\right)}^{2} = 64 \text{ is the equation}$

Expanding the brackets, rearranging the terms and equating to zero gives another form of the equation.

$\textcolor{red}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{{x}^{2} + {y}^{2} + 2 g x + 2 f y + c = 0} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

$\Rightarrow {\left(x + 8\right)}^{2} + {\left(y - 13\right)}^{2} = 64 \text{ becomes }$

${x}^{2} + 16 x + 64 + {y}^{2} - 26 y + 169 - 64 = 0$

$\Rightarrow {x}^{2} + {y}^{2} + 16 x - 26 y + 169 = 0 \text{ is another form }$