# How do you find the center and radius of the circle x^2+y^2+4x-8y+4=0?

##### 1 Answer
Jul 17, 2016

centre = (-2 ,4) , radius = 4

#### Explanation:

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

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

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

${x}^{2} + {y}^{2} + 4 x - 8 y + 4 = 0 \text{ is in this form}$

To find the centre and radius, we require to identify g , f and c

By comparing the coefficients of 'like terms' in the given equation with the general form.

2g = 4 → g = 2 , 2f = -8 → f = -4 and c = 4

$\Rightarrow \text{ centre} = \left(- g , - f\right) = \left(- 2 , 4\right)$

and radius $= \sqrt{{2}^{2} + {\left(- 4\right)}^{2} - 4} = \sqrt{4 + 16 - 4} = 4$

Alternatively Use the method of $\textcolor{b l u e}{\text{completing the square}}$

Add (1/2"coefficient of x/y terms")^2" to both sides"

(x^2+4xcolor(red)"+4")+(y^2-8ycolor(red)"+16")=color(red)"4+16"-4

$\Rightarrow {\left(x + 2\right)}^{2} + {\left(y - 4\right)}^{2} = 16$

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.

By comparison of equation with standard form.

a = -2 , b = 4 and r = 4

Thus centre = (-2 ,4) and radius = 4