Check if a circle is contained in another circle

I don't know about python but the math is simple. See the below picture

To check if circle 2 inside circle 1,

compute d 
    d = sqrt( (x2-x1)^2 + (y2-y1)^2 );
get c2 and c1
if c1 > ( d + c2 ) 
   circle 2 inside circle 1
else
   circle 2 not inside circle 1

If you want strict containment, that means that the absolute value of the difference of radii will be less than the distance between centers. You can exploit that in order to avoid taking square root (because squares of two positive numbers will have the same order as the numbers themselves):

def contains(self,circle):
    distance_squared = (circle.get_center()[0]-self.get_center()[0])**2 + (circle.get_center()[1] - self.get_center()[1])**2
    difference_squared = (self.get_radius() - circle.get_radius())**2
    return (difference_squared < distance_squared) and (self.get_radius() > circle.get_radius())

Btw, just as a style note, there is no need to write getters and setters in Python. You can just have fields and if you need to modify how they are accessed, you can override it later on (without effecting any of the classes which access them).

Making this easy from the earliest versions (maybe even from the start) was one of the reasons Python was so appealing and managed to take off. Python code tends to be very short because of this. So you don't lose sight of the forest for the trees.

You have distance_2 = distance + circle.get_radius(), so distance_2 will always be higher than distance and distance_2 > distance will always be true.

CODE ACCORDING TO THE IMAGE IN THE ACCEPTED ANSWER (without getters, as it's more readable):

import math
def contains(self, circle):
  d = math.sqrt(
        (circle.center[0] - self.center[0]) ** 2 +
        (circle.center[1] - self.center[1]) ** 2)
  return self.radius  > (d + circle.radius)

I used it, and it works. In the following plot, you can see how the circles completely contained in the red big one are painted in green, and the others in black: