# How do you write the equation of the circle in standard form x^2+y^2-6x+4y-3+0?

Dec 15, 2015

${\left(x - 3\right)}^{2} + {\left(y + 2\right)}^{2} = {4}^{2}$
$\left({x}^{2} - 2 \cdot 3 \cdot x + {3}^{2}\right) - {3}^{2} +$
$+ \left({y}^{2} + 2 \cdot 2 \cdot y + {2}^{2}\right) - {2}^{2} - 3 = 0$
$R i g h t a r r o w {\left(x - 3\right)}^{2} + {\left(y + 2\right)}^{2} = 9 + 4 + 3$