Solution. Let $C M=x, C K=y$. Triangles $A B C$ and $K M C$ are similar, since $\angle C M K=\pi-\angle A M K=\angle A B C$. Then $A C=y \sqrt{3}, B C=x \sqrt{3}$ and the area of $\triangle A B C$ is $\frac{3}{2} x y \sin \angle C$. From the fact that $A C \cdot M C$ equals the square of the length of the tangent from point $C$, we get $x y \sqrt{3}=b^{2}-a^{2}$. By the Law of Sines, $\sin \angle A M B=\frac{A B}{2 a}=\frac{\sqrt{3}}{2}$, and since the center of the circle is inside the triangle, $\angle A M B=60^{\circ}$. Further, $\sin \angle M B K=\frac{M K}{2 a}=\frac{1}{2}, \angle M B K=30^{\circ}$ and $\angle C=\angle A M B-\angle M B K=30^{\circ}$. The area of $\triangle A B C$ is $\frac{1}{2} A C \cdot B C \cdot \sin \angle C=\frac{3 x y}{4}=\frac{\sqrt{3}}{4}\left(b^{2}-a^{2}\right)$.

Answers in different versions: $6 \sqrt{3} ; 8 \sqrt{3} ; 12 \sqrt{3} ; 24 \sqrt{3}$.
3. A circle of radius $a$ passes through vertices $A$ and $B$ of triangle $ABC$ and intersects sides $AC$ and $BC$ at points $M$ and $K$ respectively. Find the area of triangle $ABC$, if $AB = a \sqrt{3}$, $MK = a$, and the center of the circle is inside triangle $ABC$ at a distance $b$ from point $C$. In different versions, $a$ and $b$ were different.