You also need to encode the painted areas somehow. They are not only intersections on K shapes, but sometimes exclusions as well (like (A^B)/C). Two ways come to mind:(1) Listing closed curves by vertices. Each vertex of a painted area is an intersection of two or more circles, and delimits a section of a circle. So the section of circles that enclose a circle can be encoded each by the union of:
(1.1) A circle (index);
(1.2) A 2nd circle (index) that intersects the 1st on a first point;
(1.3) A bit identifying the (first) intersection (because there may be 2 possible);
(1.4) A 3rd circle (index) that intersects the 1st on a second point;
(1.5) A bit identifying the (second) intersection.
Note the base circle would be the first intersection of a subsequent section of this closed curve, and the 3rd circle would be the subsequent base circle. So 1/2/3 won't be necessary for subsequent curves. So only (K+2) indices + (K+1) bits are necessary for this encoding.
Total ~K log2(K)+K bits. I hypothesize (left to the reader :)) a closed curve should contain at most 2x13 points. There can be at most 2^13 distinct regions however, so each figure (Animal) can be encoded with less than that many curves per figure. So each figure (Animal) can be encoded with less than 2^13 x 26x(5+1) bits =~ 1.3Mbit.
But that's mostly pathological cases, if each Animal must be a fully connected area, then that might reduce (hypothesis above) to at most only
26x(5+1) bits = 156 bits, or 20 bytes!
I left out a problem which area shapes encoded within each other (like eyes). In that case you need at most another 156 bits per inner cutout shape.
(2) Alternatively, you could use boolean operations to encode each shape. Also left as a fun problem :)