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find the equation of the circle with a diameter that has endpoints located at (2,8) and (2,-4).
a: (x-2)^2+(y-2)^2=6
b: (x-2)^2+(y-2)^2=36
c: (x-2)^2+(y+2)^2=36
d: (x-2)^2+(y+4)^2=36

The standard equation of a circle is:

(x-h)^2+(y-k)^2=r^2, where (h,k) is the center of the circle and r is the radius.

We are given two endpoints of the diameter, so the midpoint of these points is the center of the circle.  (2,8) and (2, -4)  so the midpoint of the circle is:

((2+2)/2, (8-4)/2)

(2,2)

The radius is just half the length of the diameter.  The length of the diameter is 8--4=12 so the radius is 6 units, so our circle equation is:

(x-2)^2+(y-2)^2=36