# How do you find the center and radius for x^2+y^2-6x-6y+14=0?

Aug 7, 2018

The center is $= \left(3 , 3\right)$ and the radius is $= 2$

#### Explanation:

The general equation of a circle, center $C = \left(a , b\right)$ and radius $= r$ is

${\left(x - a\right)}^{2} + {\left(y - b\right)}^{2} = {r}^{2}$

Here, we have

${x}^{2} + {y}^{2} - 6 x - 6 y + 14 = 0$

Rearrange the equation and complete the square

${x}^{2} - 6 x + {y}^{2} - 6 y = - 14$

${x}^{2} - 6 x + 9 + {y}^{2} - 6 y + 9 = - 14 + 9 + 9$

${\left(x - 3\right)}^{2} + {\left(y - 3\right)}^{2} = 4 = {2}^{2}$

The center is $= \left(3 , 3\right)$ and the radius is $= 2$

See the graph below.

graph{(x^2+y^2-6x-6y+14)=0 [-3.16, 12.64, -0.57, 7.33]}

Aug 7, 2018

$\text{centre "=(3,3)," radius} = 2$

#### Explanation:

$\text{the equation of a circle in "color(blue)"standard form}$ is.

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

$\text{where "(a,b)" are the coordinates of the centre and r}$
$\text{is the radius}$

$\text{to obtain this form "color(blue)"complete the square}$
$\text{on both the x and y terms}$

${x}^{2} - 6 x + {y}^{2} - 6 x = - 14$

${x}^{2} + 2 \left(- 3\right) x \textcolor{red}{+ 9} + {y}^{2} + 2 \left(- 3\right) y \textcolor{m a \ge n t a}{+ 9} = - 14 \textcolor{red}{+ 9} \textcolor{m a \ge n t a}{+ 9}$

${\left(x - 3\right)}^{2} + {\left(y - 3\right)}^{2} = 4 \leftarrow \textcolor{b l u e}{\text{in standard form}}$

$\text{centre "=(3,3)" and } r = \sqrt{4} = 2$