Question 582551
{{{(x-h)^2+(y-k)^2=r^2}}} Start with the general equation of a circle.



{{{(x-5)^2+(y--3)^2=r^2}}} Plug in {{{h=5}}} and {{{k=-3}}} (since the center is the point (h,k) ).



{{{(-1-5)^2+(-4--3)^2=r^2}}} Plug in {{{x=-1}}} and {{{y=-4}}} (this is the point that lies on the circle, which is in the form (x,y) ).



{{{(-6)^2+(-1)^2=r^2}}} Combine like terms.



{{{36+1=r^2}}} Square each term.



{{{37=r^2}}} Add.



So because  {{{h=5}}}, {{{k=-3}}}, and {{{r^2=37}}}, this means that the equation of the circle with center (5,-3) that goes through the point (-1,-4) is {{{(x-5)^2+(y+3)^2=37}}}.