How do you write the equation in standard form for a circle given just the end-point and diameter?

There are only 2 things you need to write the equation of a circle: the coordinates of its center, and its radius.  Assuming that a line connecting those 2 points is the diameter, you've got everything you need right there.  First, we get the center - it's going to be the mid-point of the line connecting those 2 points.  You can get that simply by averaging the x-coordinates and the y-coordinates from the 2 points.  For x, you've got -3 and 5, so the average is 1.  For y, you've got 4 and -6, so the average is -1.  This gives us the coordinates of the center of the circle as (1,-1).  Next you need the radius, which is half of the diameter.  The diameter is found simply using the Pythagorean theorem and it's the square root of 164, so the radius is (root(164))/2 (it's ugly, but since we actually use the radius squared in the equation it ends up being ok).

Now that we've got the center of the circle and the radius, writing the equation of the circle is a piece of cake.  The equation is (x minus x-coordinate)^2 + (y minus y-coordinate)^2 = radius^2.  So plugging in our numbers, we've got (x - 1)^2 + (y - -1)^2 = ((root(164))/2)^2.  We can simplify in 2 ways: first, y - -1 is the same as saying y + 1, and second is that ((root(164))/2)^2 is equal to 164/4, which equals 41.  So the final, simplified equation should be:
(x-1)^2 + (y+1)^2 = 41

Let A = (-3, 4) and let B = (5, -6).

I'm hoping that by end-point you mean that AB is a diameter.

If that is the case, then the center of the circle is at

X = (h, k) = (A + B)/2 = (1, -1).

The radius is given by
r^2 = XA^2 = XB^2 = (1-5)^2 + (-1+6)^2 = 41

The equation of the circle is then (x - h)^2 + (y - k)^2 = r^2

(x - 1)^2 + (y + 1)^2 = 41