# How do you find the center and radius of the circle x^2 + y^2 - 6x + 8y = 0?

May 26, 2018

$\text{centre "=(3,-4)" and radius } = 5$

#### Explanation:

$\text{the equation of a circle in standard form is}$

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

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

$\text{use the method of "color(blue)"completing the square}$
$\text{on both the x and y terms}$

${x}^{2} - 6 x + {y}^{2} + 8 y = 0$

${x}^{2} + 2 \left(- 3\right) x \textcolor{red}{+ 9} + {y}^{2} + 2 \left(4\right) y \textcolor{m a \ge n t a}{+ 16} = 0 \textcolor{red}{+ 9} \textcolor{m a \ge n t a}{+ 16}$

${\left(x - 3\right)}^{2} + {\left(y + 4\right)}^{2} = 25 \leftarrow \textcolor{b l u e}{\text{in standard form}}$

$\text{centre "=(3,-4)" and radius } = \sqrt{25} = 5$