Find the equation of a circle with the centre on the x-axis. The circles passes through points (2,2) & (-4,-4)?

I just can't seem to solve this problem, so far i've done this:

Find the equation of a circle with the centre on the x-axis. The circles passes through points (2,2) & (-4,-4)

Since it's centre is on the x axis, the y coordinate of the center is 0.

Let's name the points A and B:

A(2, 2)

B(-4, -4)

So for A:

q=0 therefore,

(2-p)^2+(2-0)^2 = r^2

4 - 4p +p^2 + 4 = r^2

p^2 - 4p + 8 = r^2

And for B:

(-4-p)^2+(-4-0)^2 = r^2

16 + 8p + p^2 +16 = r^2

p^2 + 8p + 32 = r^2

Now since both of these points can be found on the circle, we know that the radii are the same, so we can equal them.

p^2 - 4p + 8 = p^2 + 8p + 32

p^2 + 8p + 32 - p^2 + 4p - 8 = 0

12p = -24

p = -2

So

p = -2

q = 0

Therefore the equation of the circle is

(x + 2)^2 + y^2 = r^2

Now the thing i don't get is: how do you get the radius?