College algebra, diameter of a circle problem please help

Given that the diameter of a circle has endpoints A(- 9 , 4) and B(-1,-2), find:
78) the distance from A to B
i got 10 for this

79) the midpoint of the line segment connecting A and B
i got (-5,1)

80) the equation of a circle with line segment AB as its diameter
this is where im lost. (x+5)^2+(y-1)^2=25 is the answer. I can get x+5 and y-1, but how do you get 25?

81) the slope of AB
m= -3/4 I dont know how to get this either.

Thanks for the help

78
The generic formula for distance (D) between two points: D^2= (x1-x2)^2 + (y1-y2)^2
step 1) sub in known values and simplifly
D^2= (-9--1)^2 + (4--2)^2 = 8^2 + 6^2 = 64+36=100 (be careful of the negatives)
D^2=100 (take the SQ root of both sides)
D=10 Answer

#79 The generic formula for Midpoint(x) = (x1+x2)/2, Midpoint (y) = (y1+y2)/2
step 1) sub in known values and simplifly

Midpoint (x) = ((-9)+(-1))/2 = -10/2 =-5
Midpoint (y) = (4+(-2))/2 = 2/2=1

Midpoint (-5, 1) answer

#80 The generic formula for a circle (x-a)^2+(y-b)^2=R^2, where "a" is the X value of the center point of the circle, "b" is the Y value of the center point of the circle, and R is the radius of the circle.

step 1) find a, b, R (hint you just solved these in #78 and #79)
The midpoint of the line (-5, 1) is the center of the circle.
the radius (R) is one half the Diameter and the Diameter of the circle is the same as the distance of the line.
Therefore:
a=-5
b=1
R=D/2=10/2=5
Step 2) sub in known values

(x--5)^2 + (y-1)^2 = 5^2

Step 3) simplify
(x+5)^2 + (y-1)^2 = 25 Answer

#81 the generic formula for slope (m) is : m=rise/run or m=(y2-y1)/(x2-x1)
Step 1) sub in known veriables and simplify:

m = (-2-4)/(-1--9) = -6/8 = -3/4 (be careful about the negatives)
m= -3/4 answer

78 and 79 is correct
for 80) the equation of a circle is (x-h)^2 + (y-k)^2 = r^2, and since ur radius is 5, r^2 = 25
and for 81) the slope formula is (y2-y1)/(x2-x1) and since A(-9,4) and B(-1,-2), (-2 - 4)/(-1 - -9) = -3/4