# How do you find the radius of the circle x^2 + y^2 - 4x + 6y - 12 = 0?

May 26, 2016

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

#### Explanation:

The circle equation can be arranged as

${\left(x - {x}_{0}\right)}^{2} + {\left(y - {y}_{0}\right)}^{2} = {r}^{2}$

in which ${x}_{0} , {y}_{0}$ are the center coordinates and $r$ the radius.
Expanding and comparing

${x}^{2} + {y}^{2} - 2 {x}_{0} x - 2 {y}_{0} y + {x}_{0}^{2} + {y}_{0}^{2} - {r}^{2} = 0$
${x}^{2} + {y}^{2} - 4 x + 6 y - 12 = 0$

so we have

$\left\{\begin{matrix}- 2 {x}_{0} = - 4 \\ - 2 {y}_{0} = 6 \\ {x}_{0}^{2} + {y}_{0}^{2} - {r}^{2} = - 12\end{matrix}\right.$

Solving for ${x}_{0} , {y}_{0} , r$ easily we obtain

${x}_{0} = 2 , {y}_{0} = - 3 , r = 5$

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