# How to find the center and radius of x^2+y^2-6x-10y=-9?

Jun 6, 2018

The centre is $\left(3 , 5\right)$ and the radius is 5

#### Explanation:

We know that the general form of a circle looks something like this:
${\left(x - h\right)}^{2} + {\left(y - k\right)}^{2} = {r}^{2}$
where $\left(h , k\right)$ is the centre and r is the radius

${x}^{2} + {y}^{2} - 6 x - 10 y = - 9$ can be solved by completing the square

$\left({x}^{2} - 6 x\right) + \left({y}^{2} - 10 y\right) = - 9$

$\left({x}^{2} - 6 x + 9\right) + \left({y}^{2} - 10 y + 25\right) = - 9 + 9 + 25$

${\left(x - 3\right)}^{2} + {\left(y - 5\right)}^{2} = 25$

Therefore, the centre is $\left(3 , 5\right)$ and the radius is 5

Jun 6, 2018

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

#### Explanation:

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

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

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

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

${x}^{2} - 6 x + {y}^{2} - 10 y = - 9$

${x}^{2} + 2 \left(- 3\right) x \textcolor{red}{+ 9} + {y}^{2} + 2 \left(- 5\right) y \textcolor{m a \ge n t a}{+ 25} = - 9 \textcolor{red}{+ 9} \textcolor{m a \ge n t a}{+ 25}$

${\left(x - 3\right)}^{2} + {\left(y - 5\right)}^{2} = 25 \leftarrow \textcolor{b l u e}{\text{in standard form}}$

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