Very interesting problem, how do you define 2 unit circles, are they the same or like do they have integer radius? If 2 circles cover, there have to be a relation between the radical axis of the 2 circles and the sides of the right angle triangle

let the 3 point of the triangle be ABC, I feel the idea is to consider 4 cases where: Case 1: Non of ABC lie on the unit circle's circumcircle Case 2: One of ABC lie on a the circumcircle of a unit circle Case 3: Two of ABC lie on a the circumcircle of a unit circle Case 4: Three of ABC lie on a the circumcircle of a unit circle After considering these 4 cases establish an inequality using the 3 sides of triangle ABC, this idea might help find the maximum area of the right triangle

Answer: $\frac{3\sqrt{3}}{2}$ The construction is easy, just consider half of an equilateral triangle with circumradius $2$. Let the right triangle be $ABC$ with right angle at $\angle A$. Let the two circles meet at $D$ and $E$ and let $DE$ meet the perimeter of $ABC$ at $X$ and $Y$. Segment $XY$ must divide $ABC$ into two pieces that can be covered by a unit circle. If $X$ and $Y$ lie on $AB$ and $AC$ then $BCXY$ is coverable by a unit circle so $BC<2\rightarrow [ABC]\leq 1$. No WLOG assume that $X\in AB$ and $Y\in BC$. Then one establishes that $BX\leq 2$ and $CX\leq 2$. Let $AX=x$ and $AC=y$ then it suffices to show that $$x^2+y^2\leq 4 \Rightarrow y(x+2)\leq 3\sqrt{3}$$But this follows from $$y(x+2)=\sqrt{y^2(x+2)^2}\leq \sqrt{(4-x^2)(x+2)^2}\leq \left(\frac{(4-x^2)+(x+2)+(x+2)}{3}\right)^{\frac{3}{2}}=\left(\frac{9-(x-1)^2}{3}\right) ^{\frac{3}{2}}\leq 3\sqrt{3}$$