Following from Benoît Kloeckner's comment above,

Place the points at $A=(0,0)$ at the origin, $B=(c,0)$ on the x-axis with the distance $|AB|=c$, and $C=(x,y)$ in the first quadrant, where we now want to satisfy $|AC|=b$ and $|BC|=a$.

Simple application of the Pythagorean theorem leads to

$x^2+y^2 = b^2$ and $(x-c)^2+y^2 = a^2$

as the two constraints to be applied.

Expanding and subtracting the two equations:

$x^2-2cx+c^2+y^2=a^2$ and $x^2 + y^2 =b^2$

$2cx-c^2=b^2-a^2$

$2cx = (b^2-a^2+c^2)$

$x = \frac{b^2-a^2+c^2}{2c}$

Now you can define $y$ in terms of $x$.

Simply scale the points $A=(0,0), B=(0,c)$, and $C=(x,y)$ by their respective $(u,v,w)$ barycentric coordinates to get $D=(x_D,y_D)$ as a function of $a,b,c,u,v,w$, apply the Pythagorean theorem again to get the square root of $(x_d^2 + y_d^2)$. This last step shouldn't need to be spelled out for you, but $D=u\vec{A}+v\vec{B}+w\overarrow{C}

Following from Benoît Kloeckner's comment above,

Place the points at $A=(0,0)$ at the origin, $B=(c,0)$ on the x-axis with the distance $|AB|=c$, and $C=(x,y)$, where we now want to satisfy $|AC|=b$ and $|BC|=a$.

Simple application of the Pythagorean theorem leads to

$x^2+y^2 = b^2$ and $(x-c)^2+y^2 = a^2$

as the two constraints to be applied.

Expanding and subtracting the two equations:

$x^2-2cx+c^2+y^2=a^2$ and $x^2 + y^2 =b^2$

$2cx-c^2=b^2-a^2$

$2cx = (b^2-a^2+c^2)$

$x = \frac{b^2-a^2+c^2}{2c}$

Now you can define $y$ in terms of $x$.

Simply scale the points $\vec{A}=(0,0), \vec{B}=(0,c)$, and $\vec{C}=(x,y)$ by their respective $(u,v,w)$ barycentric coordinates to get $D=(x_D,y_D)$ as a function of $a,b,c,u,v,w$, apply the Pythagorean theorem again to get $d = |\vec{D}|$ = the square root of $(x_d)^2 + (y_d)^2$. This last step shouldn't need to be spelled out for you, but $\vec{D}=u\vec{A}+v\vec{B}+w\vec{C}$