Question

write the standard equation of a circle with a center at (5,-6) passing through (-9,6)

Answer 1

general form is \(x-h)^2+(y-k)^2=r^2\) in your case
\[h=5,k=-6, r^2=(5+9)^2+(-6-6)^2\]

Answer 2

So you simply plug in the x and y values of the point you are given to what you already know... (x-5)^2+(y+6)^2 ?

Answer 3

this looks good,
(x-5)^2+(y+6)^2
but don't forget the = r^2,
so,
(x-5)^2+(y+6)^2 = r^2

this form is correct for the answer except you need r to be a number not r.
plug in your x,y point and solve for r ^_^

Answer 4

yeah what @DemolisionWolf said, don't forget the \(r^2\) part