Both circles have same center: make that point C.
Let A = halfway (tangent) point of tangent line.
Let B = right end of tangent line (could be left end).
Let r = radius of inner circle.
Let r + x = radius of outer circle.
We now have right triangle ABC: AB = 12, AC = r, BC = r + x.
Per Big Pete Pythagoras: (r + x)^2 = r^2 + 12^2.
Simplified: x^2 + 2rx = 144 
(P = pi): Area outer circle = P(r + x)^2
Area inner circle = Pr^2
Let K = shaded area; then:
K = area outer - area inner
K = P(r + x)^2 - Pr^2
K/P = (r + x)^2 - r^2 ; simplified:
K/P = x^2 + 2rx 
Substitute  in :
K/P = 144
K = ~3.14 * 144 = ~452.16
What's interesting(?!) here is r can be any value;
x will decrease proportionally as r increases.
We're all here because we're not all there !