Question 636105
{{{(x-h)^2+(y-k)^2=r^2}}} Start with the general equation of a circle.



{{{(x--2)^2+(y-4)^2=r^2}}} Plug in {{{h=-2}}} and {{{k=4}}} (since the center is the point (h,k) ).



{{{(7--2)^2+(-1-4)^2=r^2}}} Plug in {{{x=7}}} and {{{y=-1}}} (this is the point that lies on the circle, which is in the form (x,y) ).



{{{(9)^2+(-5)^2=r^2}}} Combine like terms.



{{{81+25=r^2}}} Square each term.



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



So because  {{{h=-2}}}, {{{k=4}}}, and {{{r^2=106}}}, this means that the equation of the circle with center (-2,4) that goes through the point (7,-1) is 



{{{(x+2)^2+(y-4)^2=106}}}.