Acircle in the xy plane has a diameter with endpoints (-2,3) and (8,3). if the point (0,b) lies on the circle and b > 0, what is the value of b? explain plz

Diameter endpoints: A(-2,3), B(8,3) => centre is at midpoint of the endpoints
Centre O( (-2+8)/2, (3+3)/2) = O(3,3)
Radius = distance AO = 3-(-2)=5   [because y-coordinates are both 3]
The standard equation of a circle is 
(x-xo)^2+(y-yo)^2=r^2

Substituting O(xo,yo)=O(3,3), and r=5
Circle C : (x-3)^2+(y-3)^2=5^2

Line passing through (0,b) is
L :  x=0

Substitute L in C
(0-3)^2+(y-3)^2 = 25
=>
(y-3)^2=25-3^2=16
=>
y-3=+4 or y-3=-4
=>
y=7 or y=-1

We are given that b>0, which means we reject the solution y=-1
so y=7, or b=7, and the point (0,b) = (0,7)

The very first thing to do is to find the diameter's length. That just means find the length between the endpoints.

You also need the center of the circle.

Step One
Find the diameter.
r^2 = (x2 - x1)^2 + (y2 - y1)^2
x2 = 8
x1 = - 2
y2 = 3
y1 = 3

d^2 = (8 - - 2)^2 + (3 - 3)^2
d^2 = 10^2    + 0
d^2 = 10^2
Take the square root of both sides.
d = 10    

Step One A
Find the radius.
r = d / 2
r = 10 / 2
r = 5

Step Two
Find the center.
That is found averaging the endpoints.
y_c = (3 + 3)/2 = 3
x_c = (8 - 2)/2 = 3

So the center is (3,3)

Step Three
Find the equation of the circle.
(x - 3)^2 + (y - 3)^2 = 5^2   The center and the radius are all you need to describe the circle.

Step Four
Substitute (0,b) into the equation.
(0 - 3)^2 + (b - 3)^2 = 5^2
9 + (b^2 - 6b + 9) = 25

b^2 - 6b + 18 = 25    Subtract 100 from both sides.
b^2 - 6b + 18 - 25 = 0
b^2 - 6b - 7 = 0

Step Five
Factor the equation from step 4
(b - 7)(b + 1) = 0                The given in the question eliminates b + 1 which will make b <  0.

b - 7 = 0
b = 7             <<<< answer