# How do you find the equation for a circle that passes through (2,-3) with center (1,3)?

${\left(x - 1\right)}^{2} + {\left(y - 3\right)}^{2} = 13$
Any circle with centre $\left(a , b\right)$ and radius $r = \sqrt{{x}^{2} + {y}^{2}}$ has equation ${\left(x - a\right)}^{2} + {\left(y - b\right)}^{2} = {r}^{2}$.
So in this case, $r = \sqrt{{2}^{2} + {3}^{2}} = \sqrt{13}$.
So equation is ${\left(x - 1\right)}^{2} + {\left(y - 3\right)}^{2} = 13$.