If the tangent cuts $AX,AY$ at $M,N,$ then the area $[AMN]$ is minimum, if and only if $\angle MNA=60^{\circ}.$ This follows from the well known property: Among all triangles with fixed incircle, the equilateral one has the least area. Proof: Using standard notations for the scalene triangle $\triangle ABC,$ by AM-GM we have: $\sqrt[3]{(s-a)(s-b)(s-c)} \le \frac{(s-a)+(s-b)+(s-c)}{3}=\frac{s}{3} \Longrightarrow$ $s^4 \ge 27s(s-a)(s-b)(s-c) \Longrightarrow \ s^2 \ge 3\sqrt{3} [ABC] \Longrightarrow [ABC] \ge 3\sqrt{3}r^2$ The right side of the latter inequality is indeed the area of an equilateral triangle with inradius $r.$