Answer
Center: $(2,-3)$
Radius: $6$ units
Work Step by Step
Grouping the $x$-terms and the $y$-terms, the given equation, $ x^2+y^2-4x+6y-23=0 ,$ is equivalent to \begin{align*} (x^2-4x)+(y^2+6y)-23+23&=0+23 \\ (x^2-4x)+(y^2+6y)&=23 .\end{align*} Completing the square in each group results to \begin{align*} \left(x^2-4x+\left(\dfrac{-4}{2}\right)^2\right)+\left(y^2+6y+\left(\dfrac{6}{2}\right)^2\right)&=23+\left(\dfrac{-4}{2}\right)^2+\left(\dfrac{6}{2}\right)^2 \\\\ \left(x^2-4x+4\right)+\left(y^2+6y+9\right)&=23+4+9 \\\\ \left(x^2-4x+4\right)+\left(y^2+6y+9\right)&=36 .\end{align*} Using $a^2\pm2ab+b^2=(a\pm b)^2$ or the factoring of perfect square trinomials, the equation above is equivalent to \begin{align*}\require{cancel} \left(x-2\right)^2+\left(y+3\right)^2&=36 \\ \left(x-2\right)^2+\left(y-(-3)\right)^2&=36 .\end{align*} Using $(x-h)^2+(y-k)^2=r^2$ or the Standard Form of the equation of circles, the point $(h,k)$ is the center and $r$ is the radius. Hence, in the equation $\left(x-2\right)^2+\left(y-(-3)\right)^2=36$, the center is $(2,-3)$ and the radius is $6$ units.
