# How do you write the standard equation of the circle with center (-4, 3) and a radius of 5?

${x}^{2} + {y}^{2} + 8 x - 6 y = 0 \text{ " }$standard form
${\left(x + 4\right)}^{2} + {\left(y - 3\right)}^{2} = 25 \text{ " }$center-radius form
${\left(x - - 4\right)}^{2} + {\left(y - 3\right)}^{2} = {5}^{2}$
${x}^{2} + 8 x + 16 + {y}^{2} - 6 y + 9 = 25$
${x}^{2} + {y}^{2} + 8 x - 6 y = 0$