# How do you find the center and radius of the circle x^2 + y^2 + 4x - 4y - 1 = 0?

Aug 5, 2016

centre = (-2 ,2) , radius = 3

#### Explanation:

The $\textcolor{b l u e}{\text{general equation of a circle}}$ is

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

with centre = $\textcolor{red}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{\left(\left(- g , - f\right)\right)} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

and radius (r) = $\textcolor{red}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{\sqrt{{g}^{2} + {f}^{2} - c}} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

the equation ${x}^{2} + {y}^{2} + 4 x - 4 y - 1 = 0 \text{ is in this form}$

and by comparing the coefficients of like terms we get the values for g ,f and c.

$2 g = 4 \Rightarrow g = 2 , 2 f = - 4 \Rightarrow f = - 2 \text{ and } c = - 1$

$\Rightarrow \text{ centre} = \left(- g , - f\right) = \left(- 2 , 2\right)$

and r = $\sqrt{{g}^{2} + {f}^{2} - c} = \sqrt{{2}^{2} + {\left(- 2\right)}^{2} - \left(- 1\right)} = \sqrt{9} = 3$
$\textcolor{b l u e}{\text{---------------------------------------------------------------------------}}$

Alternatively the method of $\textcolor{b l u e}{\text{completing the square}}$ can be used to obtain the equation.

$\left({x}^{2} + 4 x \textcolor{red}{+ 4}\right) + \left({y}^{2} - 4 y \textcolor{red}{+ 4}\right) - 1 = \textcolor{red}{4} + \textcolor{red}{4}$

$\Rightarrow {\left(x + 2\right)}^{2} + {\left(y - 2\right)}^{2} = 9$

The$\textcolor{b l u e}{\text{ standard form of the equation of a circle}}$ 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 coordinates of the centre and r, the radius.

$\Rightarrow \text{centre"=(-2,2)" and } r = 3$