# How do you rewrite the following equation in the center-radius form of the equation of a circle x^2 - 4x + y^2 - 4y + 4 = 0?

Jun 26, 2016

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

#### Explanation:

The general form of the equation of a circle is

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

centre = (-g ,-f) and radius (r)=$\sqrt{{g}^{2} + {f}^{2} - c}$

If we compare like terms in the given equation to the general form.

then 2g = - 4 → g = -2 , 2f = -4 → f = -2 and c = 4

hence centre = (2 , 2) and r $= \sqrt{{\left(- 2\right)}^{2} + {\left(- 2\right)}^{2} - 4} = 2$

The equation of the circle in $\textcolor{b l u e}{\text{centre-radius form}}$ 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.

substitute a = 2 , b =2 and r =2 into this equation

$\Rightarrow {\left(x - 2\right)}^{2} + {\left(y - 2\right)}^{2} = 4 \text{ is the equation}$