smc

In this solution, let the brackets denote areas. We place the diagram in the coordinate plane: Let $A=(12,0),B=(0,5),$ and $C=(0,0).$ Since $△ABC$ is a right triangle with $\angle ACB=90^{\circ },$ its circumcenter is the midpoint of $\overline{AB},$ from which $O=\left(6,\frac{5}{2}\right).$ Note that the circumradius of $△ABC$ is $\frac{13}{2}.$ Let $s$ denote the semiperimeter of $△ABC.$ The inradius of $△ABC$ is $\frac{[ABC]}{s}=\frac{30}{15}=2,$ from which $I=(2,2).$ Since $\odot M$ is also tangent to both coordinate axes, its center is at $M=(a,a)$ and its radius is $a$ for some positive number $a.$ Let $P$ be the point of tangency of $\odot O$ and $\odot M.$ As $\overline{OP}$ and $\overline{MP}$ are both perpendicular to the common tangent line at $P,$ we conclude that $O,M,$ and $P$ are collinear. It follows that $OM=OP-MP,$ or $$\sqrt{(a-6{)}^2+{\left(a-\frac{5}{2}\right)}^2}=\frac{13}{2}-a.$$ Solving this equation, we have $a=4,$ from which $M=(4,4).$ Finally, we apply the Shoelace Theorem to $△MOI:$ $$[MOI]=\frac{1}{2}\left|\left(4\cdot \frac{5}{2}+6\cdot 2+2\cdot 4\right)-\left(4\cdot 6+\frac{5}{2}\cdot 2+2\cdot 4\right)\right|=\boxed{\frac{7}{2}}.$$ Remark Alternatively, we can use $\overline{MI}$ as the base and the distance from $O$ to $\overleftrightarrow{MI}$ as the height for $△MOI:$ * By the Distance Formula, we have $MI=2\sqrt{2}.$ * The equation of $\overleftrightarrow{MI}$ is $x-y+0=0,$ so the distance from $O$ to $\overleftrightarrow{MI}$ is $h_O=\frac{|1\cdot 6+(-1)\cdot \frac{5}{2}+0|}{\sqrt{1^2+(-1{)}^2}}=\frac{7}{4}\sqrt{2}.$ Therefore, we get $$[MOI]=\frac{1}{2}\cdot MI\cdot h_O=\frac{7}{2}.$$