# How do you find the center and radius of the circle x^2+y^2-2x+6y-26=0?

Jun 24, 2018

$\text{centre "=(1,-3)" radius } = 6$

#### Explanation:

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

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

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

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

${x}^{2} - 2 x + {y}^{2} + 6 y = 26$

${x}^{2} + 2 \left(- 1\right) x \textcolor{red}{+ 1} + {y}^{2} + 2 \left(3\right) y \textcolor{m a \ge n t a}{+ 9} = 26 \textcolor{red}{+ 1} \textcolor{m a \ge n t a}{+ 9}$

${\left(x - 1\right)}^{2} + {\left(y + 3\right)}^{2} = 36 \leftarrow \textcolor{b l u e}{\text{in standard form}}$

$\text{with centre "=(1,-3)" radius } = \sqrt{36} = 6$
graph{((x-1)^2+(y+3)^2-36)=0 [-20, 20, -10, 10]}