Annulus Area

Let the big radius, small radius and area of the annulus be $R$ , $r$ and $A$ .

Then, $A = \pi (R^2 - r^2) = x\pi \quad \Rightarrow x = R^2 - r^2$ .

Note that line $AB$ is tangent to the inner circle. Let the perpendicular from $O$ meets $AB$ at $P$ .

Then $OP = r$ , $OA = R$ and $x = R^2 - r^2 = \overline { AP } ^2 = 9^2 = \boxed {81}$

$2a^{2} = 18^{2} , a = 9\sqrt{2}$

Now ,

$9^{2} + b^{2} = a^{2}$ (any line passing through center divides the chord into 2 equal halves)

$b = 9$

Brown Area = area of bigger circle - area of smaller circle

$= \pi(9\sqrt{2})^{2} - \pi(9)^{2} = 81\pi$

Imagine that the brown circle grows larger. (Keep in mind that none of the information provides the actual radius/diameter of the circle.) What will happen to the little circle? It will have to grow bigger so that AB remains tangent to it.

Now imagine that the brown circle grows smaller. As it does, the little circle shrinks (again so that AB stays tangent).

Let the brown circle shrink down so that the little circle disappears. Now AB is the diameter of the brown circle, its radius is 9, and the area is 81pi.

Formula applied for such concentric circles= R^2= r^2 +(c/2)^2 Area of the brown area= π( R^2-r^2) =π(c/2)^2 (Substituting the value of R^2-r^2 ) =π(9)^2 =π81

Let the point where line AB is tangent to the small circle be c. Now angle(OCA)=90 degrees (tangent to a radius). Using another circle theorem: a radius bisects a perpendicular chord, so AC=CB=9. Let r=radius of small circle and R=radius of large circle. By Pythagoras' Theroem, r^{2} + 81 = R^{2}. The area of the brown is equal to difference in the areas of the 2 circles; therefore x pi = pi R^{2} - pi r^{2} = pi [R^{2} - r^{2} ] = 81*pi. Hence, x=81