# How do you find the center and radius of the circle x^2 + y^2 - 12x - 8y + 48 = 0?

Write this as follows

${x}^{2} + {y}^{2} - 12 x - 8 y + 48 = 0$

$\left({x}^{2} - 12 x + 36\right) + \left({y}^{2} - 8 y + 16\right) + 48 - 36 - 16 = 0$

${\left(x - 6\right)}^{2} + {\left(y - 4\right)}^{2} = 4$

${\left(x - 6\right)}^{2} + {\left(y - 4\right)}^{2} = {2}^{2}$

From this we get the center which is point $\left(6 , 4\right)$ and radius $r = 2$